# Signed circuit $6$-covers of signed $K_{4}$-minor-free graphs

You Lu Rong Luo Zhengke Miao and Cun-Quan Zhang Research & Development
Institute of Northwest Polytechnical University in Shenzhen, Shenzhen,
Guangdong 518063, China; School of Mathematics and Statistics, Northwestern
Polytechnical University, Xi’an, Shaanxi 710129, China. Email:
<EMAIL_ADDRESS>Partially supported by NSFC (No. 12271438), Guangdong Basic
and Applied Basic Research Foundation (No. 2023A1515012340) and Natural
Science Foundation of Qinghai Province (No. 2022-ZJ-753).Department of
Mathematics, West Virginia University, Morgantown, WV 26505, USA. Email:
<EMAIL_ADDRESS>Partially supported by a grant from Simons Foundation (No.
839830)Research Institute of Mathematical Science and School of Mathematics
and Statistics, Jiangsu Normal University, Xuzhou, Jiangsu 221116, China.
Email<EMAIL_ADDRESS>Partially supported by NSFC (No.
11971205)Department of Mathematics, West Virginia University, Morgantown, WV
26505, USA. Email<EMAIL_ADDRESS>Partially supported by an NSF grant
DMS-1700218

###### Abstract

Bermond, Jackson and Jaeger [J. Combin. Theory Ser. B 35 (1983): 297-308]
proved that every bridgeless ordinary graph $G$ has a circuit $4$-cover and
Fan [J. Combin. Theory Ser. B 54 (1992): 113-122] showed that $G$ has a
circuit $6$-cover which together implies that $G$ has a circuit $k$-cover for
every even integer $k\geq 4$. The only left case when $k=2$ is the well-know
circuit double cover conjecture. For signed circuit $k$-cover of signed
graphs, it is known that for every integer $k\leq 5$, there are infinitely
many coverable signed graphs without signed circuit $k$-cover and there are
signed eulerian graphs that admit nowhere-zero $2$-flow but don’t admit a
signed circuit $1$-cover. Fan conjectured that every coverable signed graph
has a signed circuit $6$-cover. This conjecture was verified only for signed
eulerian graphs and for signed graphs whose bridgeless-blocks are eulerian. In
this paper, we prove that this conjecture holds for signed $K_{4}$-minor-free
graphs. The $6$-cover is best possible for signed $K_{4}$-minor-free graphs.

##  1 Introduction

Graphs or signed graphs considered in this paper are finite and may have
multiple edges or loops. For terminology and notations not defined here we
follows [6, 10, 21, 27].

A _signed graph_ is a graph $G$ with a mapping $\sigma:E(G)\mapsto\\{1,-1\\}$.
The mapping $\sigma$, called _signature_ , is sometimes implicit in the
notation of a signed graph and will be specified when needed. An edge $e$ is
positive if $\sigma(e)=1$, and otherwise it is negative. An ordinary graph is
a signed graph without negative edges and a circuit is a connected $2$-regular
graph. A circuit in a signed graph is balanced if it has an even number of
negative edges and otherwise it is unbalanced. A _signed circuit_ is either a
balanced circuit or a barbell, the union of two unbalanced circuits and a
(possibly trivial) path (called the barbell-path) that meets the circuits only
at ends. A barbell is called a short barbell if its barbell-path is trivial,
and a long barbell otherwise. The edges of a signed circuit in a signed graph
correspond to a minimal dependent set in the signed graphic matroid (see
[30]).

Let $G$ be a signed graph. A family $\mathcal{F}$ of signed circuits of $G$ is
called a signed circuit cover of $G$ if every edge is contained in some member
of $\mathcal{F}$ and is called a signed circuit $k$-cover if each edge is
contained in precisely $k$ members of $\mathcal{F}$. A signed graph is
coverable if it has a signed circuit cover. Given a coverable signed graph
$G$, the minimum length of a signed circuit cover of $G$ is denoted by
$SSC(G)$.

Note that there is no unbalanced circuit and thus no barbell in an ordinary
graph. The circuit cover of ordinary graphs is closely related to some
mainstream areas in graph theory, such as, Tutte’s integer flow theory [1, 5,
13, 16, 19, 25, 31], Fulkerson conjecture [14], snarks and graph minors [2,
17]. Thus the circuit cover of ordinary graphs has been studied extensively.

It is proved by Bermond, Jackson and Jaeger [5] that every ordinary graph
admitting a nowhere-zero $4$-flow has $SCC(G)\leq\frac{4|E|}{3}$. By applying
Seymour’s $6$-flow theorem [26] or Jaeger’s $8$-flow theorem [18], Alon and
Tarsi [1], and Bermond, Jackson and Jaeger [5] proved that every bridgeless
ordinary graph $G$ has $SCC(G)\leq\frac{25|E|}{15}$. One of the most famous
open problems in this area was proposed by Alon and Tarsi [1] that every
bridgeless ordinary graph $G$ has $SCC(G)\leq\frac{21|E|}{15}$.

Bermond, Jackson and Jaeger [5] proved that every bridgeless ordinary graph
$G$ has a circuit $4$-cover and Fan [12] showed that $G$ has a circuit
$6$-cover which together implies that $G$ has a circuit $k$-cover for every
even integer $k\geq 4$. The only left case when $k=2$ is the well-known
circuit double cover conjecture.

For signed graphs, Máčajová, Raspaud, Rollová and Škoviera [23] presented the
first upper bound of $SSC(G)$. They showed that $SSC(G)\leq 11|E(G)|$ if $G$
is coverable and the upper bound was improved by Lu et al. [22] to
$\frac{14}{3}|E(G)|$. More improvements were obtained later in [8, 20, 24, 28,
29].

For $k$-cover of signed graphs, Fan [15] showed that for every integer $k\leq
5$, there are infinitely many coverable signed graphs that have no signed
circuit $k$-cover and he proposed the following conjecture.

######  Conjecture 1.1.

(Fan [15]) Every coverable signed graph has a signed circuit $6$-cover.

The conjecture was verified for signed eulerian graphs in [4] and for signed
graphs whose bridgeless-blocks are eulerian in [7].

A graph $H$ is a minor of a graph $G$ if a graph isomorphic to $H$ can be
obtained from $G$ by edge contractions, edge deletions and vertex deletions;
if not, $G$ is $H$-minor-free. The class of $K_{4}$-minor-free graphs, which
includes all series-parallel graphs and outerplanar graphs, is a very
important family of graph class and has been studied by many researchers for
various graph theory problems (for example see [11, 9]). In this paper we
study the signed circuit $k$-cover and confirm Conjecture 1.1 for signed
$K_{4}$-minor-free graphs and confirm Conjecture 1.1 for this family of signed
graphs.

######  Theorem 1.2.

Every coverable signed $K_{4}$-minor-free graph has a signed circuit
$6$-cover.

Note that every coverable signed graph $G$ with four distinct degree $3$
vertices $x_{1},x_{2},y_{1},y_{2}$ has no signed circuit $k$-cover for any
$k\in[1,5]$ if $G[\\{x_{1},x_{2}\\}]$ is a balanced $2$-circuit and
$G[\\{y_{1},y_{2}\\}]$ is an unbalanced $2$-circuit. Thus the $6$-cover in
Theorem 1.2 is tight.

Before proceeding, it is worth pointing out that the problems of flow and
signed circuit cover in signed graphs are significantly more challenging than
their counterparts in ordinary graphs. For instance, while ordinary Eulerian
graphs trivially allow for a nowhere-zero $2$-flow and a $1$-cover, signed
Eulerian graphs can have flow values of $2$, $3$, or even $4$, as shown in
[24]. Additionally, there are signed Eulerian graphs that admit nowhere-zero
$2$-flow but don’t have a $1$-cover, as demonstrated in [4]. Unlike ordinary
graphs, coverable signed graphs may have bridges. The intricate structures of
signed graphs, including barbells, bridges, and negative loops, contribute to
their heightened complexity compared to ordinary graphs. Consequently,
tackling the flow and signed circuit cover problems in signed graphs requires
extensive machinery and specialized approaches.

This paper is organized as follows. In Section 2, we introduce more notations
and terminology. Some simple cases and reduction lemmas needed in the proof of
Theorem 1.2 are presented in Section 3. In Section 4, we prove Theorem 1.2 by
contradiction.

##  2 Preliminaries

Let $G$ be a graph. A vertex $x$ is called a cut vertex of $G$ if $G-x$ has
more components than $G$. A graph is $2$-connected if it is connected and has
no cut vertex. Let $L_{x}$ represent a loop at $x$ and $L(G)$ be the set of
all loops of $G$. Let $N_{G}(x)$ and $d_{G}(x)$ denote the neighborhood and
the degree of $x$ in $G$, respectively, where each loop at $x$ contributes $2$
to $d_{G}(x)$. A $d$-vertex is a vertex with degree $d$. For two subsets
$X,Y\subseteq V(G)$ (not necessarily disjoint), denote by $E_{G}[X,Y]$ the set
of edges of $G$ with one end in $X$ and the other end in $Y$. A path with ends
$x$ and $y$ is called an $xy$-path. A cycle of length $k$ is called a
$k$-cycle.

Let $G$ be a signed graph. For an edge subset or subgraph $S$ of $G$, denote
the set of all negative edges of $S$ by $E_{N}(S)$ and define the sign of $S$
to be $\sigma(S)=\Pi_{e\in S}\sigma(e)$. A path $P$ in $G$ is positive if
$\sigma(P)=1$, and negative otherwise. The path $P$ is called a subdivided
edge of $G$ if every internal vertex of $P$ is a $2$-vertex of $G$. The
suppressed graph of $G$, denoted by $\overline{G}$, is the signed graph
obtained from $G$ by replacing each maximal subdivided edge $P$ with a single
edge $e$ and assigning $\sigma(e)=\sigma(P)$.

Given a signed graph $G$, switching at a vertex $x$ is the inversion of the
signs of all edges incident with $x$. A signed graph $G^{\prime}$ is said to
be equivalent to $G$ if $G^{\prime}$ can be obtained from $G$ via a sequence
of switchings and is denoted by $G^{\prime}\sim G$. Define the negativeness of
$G$ by $\epsilon(G)=\min\\{|E_{N}(G^{\prime})|:G^{\prime}\sim G\\}$. A signed
graph is balanced if its negativeness is $0$ and otherwise unbalanced. That
is, a balanced signed graph is equivalent to an all-positive signed graph,
i.e. an ordinary graph. It is easy to see that a signed graph is balanced if
and only if all of its circuits are balanced.

A tadpole at $x$ is the union of an $xy$-path $P$ and an unbalanced circuit
$C$ with $V(P)\cap V(C)=\\{y\\}$. The vertex $x$ is called a tail and the path
$P$ is called a tadpole-path.

######  Definition 2.1.

Let $\mathcal{F}$ be a family of signed subgraphs of a signed graph $G$. Let
$t\in[0,3]$ be an integer and $x,y$ be two distinct vertices of $G$.

  * (1)

For each $e\in E(G)$, $\mathcal{F}(e)$ denotes the number of members in
$\mathcal{F}$ containing $e$.

  * (2)

For an edge subset or a subgraph $S$ of $G$, $\mathcal{F}$ is a _signed
subgraph $k$-cover_ of $S$ if $\mathcal{F}(e)=k$ for each edge $e$ in $S$. In
particular, $\mathcal{F}$ is a _signed circuit $k$-cover_ of $G$ if every
member of $\mathcal{F}$ is a signed circuit.

  * (3)

A $\Psi_{xy}(t)$-cover is a signed subgraph $6$-cover that consists of $t$
positive $xy$-paths, $t$ negative $xy$-paths, $t$ tadpoles at $x$, $6-2t$
tadpoles at $y$, and some signed circuits.

  * (4)

Let $xy$ be an edge. A $\Psi_{xy}^{*}(2)$-cover is a $\Psi_{xy}(2)$-cover such
that for each $u\in\\{x,y\\}$, one tadpole at $u$ doesn’t contain the vertex
in $\\{x,y\\}\setminus\\{u\\}$, and the tadpole-path of the other tadpole at
$u$ contains the edge $xy$.

Signed circuit cover and flows are closely related. It is known that a signed
graph $G$ is coverable if and only if it admits a nowhere-zero $k$-flow for
some integer $k\geq 2$. Refining the results in [3], we have the following
characterization.

######  Proposition 2.2.

A connected signed graph $G$ is coverable if and only if $\epsilon(G)\neq 1$
and there is no cut edge $b$ such that $G-b$ has a balanced component.

##  3 $\Psi_{xy}(t)$-covers of two-terminal signed graphs

For two integers $n_{1}\leq n_{2}$, let $[n_{1},n_{2}]$ denote the set of
integers between $n_{1}$ and $n_{2}$. A two-terminal signed graph $H(x,y)$ is
a connected signed nonempty graph $H$ with two specified vertices, a source
terminal $x$ and a target terminal $y$, where $x=y$ if and only if $H$ is a
negative loop. For short, we abbreviate $H(x,y)$ to $H$ if the terminals are
understood from the context.

Let $H_{i}=H_{i}(x_{i},y_{i})$ be a two-terminal signed graph for each
$i\in[1,n]$. The parallel connection $\mathcal{P}(H_{1},\dots,H_{n})$ of
$H_{1},\dots,H_{n}$ is the two-terminal signed graph obtained from
$H_{1}\cup\dots\cup H_{n}$ by identifying $x_{1},\dots,x_{n}$ into a source
terminal and identifying $y_{1},\dots,y_{n}$ into a target terminal. The
series connection $\mathcal{S}(H_{1},\dots,H_{n})$ of $H_{1},\dots,H_{n}$ is
the two-terminal signed graph with source terminal $x_{1}$ and target terminal
$y_{n}$ obtained from $H_{1}\cup\dots\cup H_{n}$ by identifying $y_{i-1}$ and
$x_{i}$ for each $i\in[2,n]$. If $G$ is a series connection of
$H_{1},\dots,H_{n}$ and $n$ is maximum with this property, then we call every
$H_{i}$ a part of $G$. Let $\mathcal{B}(G)=\\{H_{1},\dots,H_{n}\\}$ be the set
of all parts of $G$. Obviously, $\mathcal{B}(G)$ can be partitioned into three
subsets as follows:

$\displaystyle\mathcal{B}_{0}(G)$
$\displaystyle=\\{H_{i}\in\mathcal{B}(G):x_{i}=y_{i}\\},$
$\displaystyle\mathcal{B}_{1}(G)$
$\displaystyle=\\{H_{i}\in\mathcal{B}(G):x_{i}\neq y_{i},|E(H_{i})|=1\\},$
$\displaystyle\mathcal{B}_{2}(G)$
$\displaystyle=\\{H_{i}\in\mathcal{B}(G):x_{i}\neq y_{i},|E(H_{i})|\geq 2\\}.$

Note that every member of $\mathcal{B}_{0}(G)$ is a negative loop and every
member of $\mathcal{B}_{1}(G)$ is a positive or negative $K_{2}$. A series
connection is shown in Fig. 1.

$H_{1}$$H_{2}$$H_{3}$$H_{4}$$H_{5}$$H_{6}$$H_{7}$ Figure 1: A series
connection $G$ with $\mathcal{B}_{0}(G)=\\{H_{2},H_{6}\\}$,
$\mathcal{B}_{1}(G)=\\{H_{1},H_{3},H_{5}\\}$ and
$\mathcal{B}_{2}(G)=\\{H_{4},H_{7}\\}$. Solid lines are positive; dotted lines
are negative.

The next lemma will be applied in the reduction.

######  Lemma 3.1.

Let $H_{i}=H_{i}(x_{i-1},x_{i})$ for each $i\in[1,n]$ and
$G=\mathcal{S}(H_{1},\dots,H_{n})$ with $n=|\mathcal{B}(G)|$ and
$|\mathcal{B}_{2}(G)|\geq 1$. Let $\theta^{*}=(1,1,-1,-1)$ if
$|\mathcal{B}_{2}(G)|=1$ and $\theta^{*}\in\\{(1,1,-1,-1),(-1,-1,-1,-1)\\}$ if
$|\mathcal{B}_{2}(G)|\geq 2$. If every $H_{i}\in\mathcal{B}_{2}(G)$ has a
$\Psi_{x_{i-1}x_{i}}(2)$-cover, then $G$ has a signed subgraph $6$-cover

$\mathcal{F}_{0}\cup
2\mathcal{B}_{0}(G)\cup\\{P_{1},P_{2},P_{3},P_{4}\\}\cup\\{T_{1},T_{2},T_{3},T_{4}\\},$

where

  * $\rhd$

$\mathcal{F}_{0}$ is a family of signed circuits;

  * $\rhd$

$P_{1},P_{2},P_{3},P_{4}$ are four $x_{0}x_{n}$-paths of $G$ and
$(\sigma(P_{1}),\sigma(P_{2}),\sigma(P_{3}),\sigma(P_{4}))=\theta^{*}$;

  * $\rhd$

$T_{1}$ and $T_{2}$ (resp., $T_{3}$ and $T_{4}$) are two tadpoles of $G$ at
$x_{0}$ (resp., $x_{n}$) whose unbalanced circuit is in the part in
$\mathcal{B}_{0}(G)\cup\mathcal{B}_{2}(G)$ with minimum (resp., maximum)
subscript.

###### Proof.

Denote $I_{j}=\\{i:H_{i}\in\mathcal{B}_{j}(G)\\}$ for each $j\in[0,2]$, and
for each $i\in I_{2}$, let

$\mathcal{F}_{i}=\mathcal{C}_{i}\cup\\{P_{i1},P_{i2},P_{i3},P_{i4}\\}\cup\\{T_{i1},T_{i2},T_{i3},T_{i4}\\},$

be an arbitrary $\Psi_{x_{i-1}x_{i}}(2)$-cover of $H_{i}$, where
$\mathcal{C}_{i}$ is a family of signed circuits, $P_{i1},P_{i2}$ (resp.,
$P_{i3},P_{i4}$) are two positive (resp., negative) $x_{i-1}x_{i}$-paths, and
$T_{i1},T_{i2}$ (resp., $T_{i3}$, $T_{i4}$) are two tadpoles at $x_{i-1}$
(resp., $x_{i}$).

Let $\mathcal{G}_{1}=\left(\cup_{i\in
I_{2}}\\{P_{i1},P_{i2},P_{i3},P_{i4}\\}\right)\cup 4\mathcal{B}_{1}(G)$. Note
that every part in $\mathcal{B}_{0}(G)$ is a negative loop and every part in
$\mathcal{B}_{1}(G)$ is a positive or negative $K_{2}$. Then $\mathcal{G}_{1}$
can be expressed as a family $\mathcal{P}$ consisting of $4$
$x_{0}x_{n}$-paths $P_{1},P_{2},P_{3},P_{4}$ such that
$(\sigma(P_{1}),\sigma(P_{2}),\sigma(P_{3}),\sigma(P_{4}))=\theta^{*}$ and
$\mathcal{G}_{1}(e)=\mathcal{P}(e)$ for each $e\in E(G)$.

Let $\mathcal{G}_{2}=\left(\cup_{i\in
I_{2}}\\{T_{i1},T_{i2},T_{i3},T_{i4}\\}\right)\cup 4\mathcal{B}_{0}(G)\cup
2\mathcal{B}_{1}(G)$. For convenience, let $T_{ij}=H_{i}$ for each $i\in
I_{0}$ and each $j\in[1,4]$ since $H_{i}$ is a tadpole at $x_{i-1}$
(=$x_{i}$), and $I_{0}\cup I_{2}=\\{i_{1},i_{2},\dots,i_{\ell}\\}$ with $0\leq
i_{1}\leq i_{2}\leq\dots\leq i_{\ell}\leq n$. For each $j\in[1,2]$, we
construct a tadpole $T_{j}$ at $x_{0}$, a tadpole $T_{j+2}$ at $x_{n}$, and
some barbells as follows:

$\displaystyle T_{j}$ $\displaystyle=(x_{0}x_{1}\cdots x_{i_{1}-1})\cup
T_{i_{1}j},$ $\displaystyle T_{j+2}$
$\displaystyle=T_{i_{\ell}(j+2)}\cup(x_{i_{\ell}}\cdots x_{n-1}x_{n}),$
$\displaystyle B_{kj}$
$\displaystyle=T_{i_{k}(j+2)}\cup(x_{i_{k}}x_{i_{k}+1}\cdots
x_{i_{k+1}-1})\cup T_{i_{k+1}j},\forall\ k\in[1,\ell-1].$

Let $\mathcal{T}=\\{T_{1},T_{2},T_{3},T_{4}\\}\mbox{ and
}\mathcal{C}=\cup_{k=1}^{\ell-1}\\{B_{k1},B_{k2}\\}$. Obviously,
$\mathcal{G}_{2}(e)=(\mathcal{T}\cup\mathcal{C})(e)$ for each $e\in E(G)$.
Therefore, $(\cup_{i\in I_{2}}\mathcal{C}_{i})\cup\mathcal{C}\cup
2\mathcal{B}_{0}(G)\cup\mathcal{P}\cup\mathcal{T}$ is a desired signed
subgraph $6$-cover of $G$. ∎

By the definition and Lemma 3.1, the following result is straightforward and
its proof is omitted.

######  Lemma 3.2.

Let $H_{i}=H_{i}(x_{i-1},x_{i})$ for each $i\in[1,n]$ and
$G=\mathcal{S}(H_{1},\dots,H_{n})$ with $n=|\mathcal{B}(G)|\geq 2$. If
$\mathcal{B}_{0}(G)=\emptyset$, $\mathcal{B}_{2}(G)\neq\emptyset$ and every
$H_{i}\in\mathcal{B}_{2}(G)$ has a $\Psi_{x_{i-1}x_{i}}(2)$-cover, then
exactly one of the following statements holds.

  * (1)

$G$ has a $\Psi_{x_{0}x_{n}}(2)$-cover in which no tadpole at $x_{0}$ (resp.,
$x_{n}$) contains $x_{n}$ (resp., $x_{0}$);

  * (2)

$\mathcal{B}_{2}(G)=\\{H_{1}\\}$ and every $\Psi_{x_{0}x_{1}}(2)$-cover of
$H_{1}$ contains a tadpole at $x_{1}$ containing $x_{0}$;

  * (3)

$\mathcal{B}_{2}(G)=\\{H_{n}\\}$ and every $\Psi_{x_{n-1}x_{n}}(2)$-cover of
$H_{n}$ contains a tadpole at $x_{n-1}$ containing $x_{n}$.

The next lemma is another reduction technique in the proof of the main result.

######  Lemma 3.3.

Let $H_{1},H_{2},H_{2}^{\prime}$ be three two-terminal signed graphs with
source terminal $x$ and target terminal $y$, where $H_{2}$ and
$H_{2}^{\prime}$ satisfy one of the following conditions.

  * (1)

$H_{2}$ has a $\Psi_{xy}(t)$-cover for each $t\in[0,3]$ in which no tadpole at
$x$ contains $y$; $H_{2}^{\prime}$ is the signed graph $D_{1}(x,y)$ in Fig. 2.

  * (2)

$H_{2}$ has a $\Psi_{xy}(2)$-cover in which no tadpole at $x$ (resp., $y$)
contains $y$ (resp., $x$); $H_{2}^{\prime}$ is the signed graph $D_{2}(x,y)$
in Fig. 2.

If $\mathcal{P}(H_{1},H_{2}^{\prime})$ has a signed circuit $6$-cover, then so
does $\mathcal{P}(H_{1},H_{2})$.

$x$$y$$e_{1}$$e_{3}$$e_{2}$

(a) $D_{1}(x,y)$

$x$$y$$e_{1}$$e_{3}$$e_{2}$$e_{4}$

(b) $D_{2}(x,y)$

Figure 2: Two two-terminal signed graphs with terminals $x$ and $y$.

###### Proof.

Denote $G=\mathcal{P}(H_{1},H_{2})$ and
$G^{\prime}=\mathcal{P}(H_{1},H_{2}^{\prime})$. Let $\mathcal{F}$ be a signed
circuit $6$-cover of $G^{\prime}$.

We only prove the case when $H_{2}$ satisfies (1) since the augment for the
other case is very similar.

As shown in Fig. 2, it follows from the structure of the signed graph
$D_{1}(x,y)$ that, for any $C\in\mathcal{F}$,

$E(C)\cap
E(H_{2}^{\prime})\in\left\\{\emptyset,\\{e_{1},e_{2}\\},\\{e_{2},e_{3}\\},\\{e_{1},e_{3}\\},\\{e_{1},e_{2},e_{3}\\}\right\\}.$

Denote by $\mathcal{F}_{1}$ (resp., $\mathcal{F}_{2}$, $\mathcal{F}_{3}$,
$\mathcal{F}_{4}$) the set of signed circuits $C\in\mathcal{F}$ with $E(C)\cap
E(H_{2}^{\prime})=\\{e_{1},e_{2}\\}$ (resp., $=\\{e_{3},e_{2}\\}$,
$=\\{e_{1},e_{3}\\}$, $=\\{e_{1},e_{2},e_{3}\\}$). Since
$\mathcal{F}(e_{1})=\mathcal{F}(e_{2})=\mathcal{F}(e_{3})=6$,

$|\mathcal{F}_{1}|+|\mathcal{F}_{3}|+|\mathcal{F}_{4}|=|\mathcal{F}_{1}|+|\mathcal{F}_{2}|+|\mathcal{F}_{4}|=|\mathcal{F}_{2}|+|\mathcal{F}_{3}|+|\mathcal{F}_{4}|=6.$

Thus there is an integer $t\in[0,3]$ such that
$|\mathcal{F}_{1}|=|\mathcal{F}_{2}|=|\mathcal{F}_{3}|=t$ and
$|\mathcal{F}_{4}|=6-2t$. Let $\mathcal{F}_{i}=\\{C_{i1},\dots,C_{it}\\}$ for
each $i\in[1,3]$ and $\mathcal{F}_{4}=\\{C_{41},\dots,C_{4(6-2t)}\\}$. On the
other hand, by assumption, $H_{2}$ has a $\Psi_{xy}(t)$-cover

$\mathcal{C}_{0}\cup\\{P_{11},\dots,P_{1t}\\}\cup\\{P_{21},\dots,P_{2t}\\}\cup\\{P_{31},\dots,P_{3t}\\}\cup\\{P_{41},\dots,P_{4(6-2t)}\\},$

where no tadpole at $x$ contains $y$, $\mathcal{C}_{0}$ is a family of signed
circuits, each $P_{1j}$ (resp., $P_{2j}$) is a positive (resp., negative)
$xy$-path, and each $P_{3j}$ (resp., $P_{4j}$) is a tadpole at $x$ (resp.,
$y$). One can easily check that the family

$\mathcal{C}_{0}\cup\left(\mathcal{F}\setminus(\cup_{i=1}^{4}\mathcal{F}_{i})\right)\cup\left(\cup_{i=1}^{4}\cup_{j=1}^{|\mathcal{F}_{i}|}\\{(C_{ij}-E(H_{2}^{\prime}))\cup
P_{ij}\\}\right)$

is a signed circuit $6$-cover of $G=\mathcal{P}(H_{1},H_{2})$. ∎

Next we introduce six small signed graphs $R_{i}$s in Fig. 3. to which some
subgraphs of a smallest counterexample is reduced. The following observation
is straightforward.

$x$$y$

(a) $R_{0}$

$x$$y$

(b) $R_{1}$

$x$$y$

(c) $R_{2}$

$x$$y$

(d) $R_{3}$

$x$$y$

(e) $R_{4}$

$x$$y$

(f) $R_{5}$

Figure 3: Six small signed graphs with two specified vertices $x$ and $y$.

######  Observation 3.4.

(1) $R_{2}$ has a $\Psi_{yx}(t)$-cover for each $t\in[0,3]$, and a
$\Psi_{xy}(2)$-cover in which exactly one tadpole at $y$ doesn’t contain $x$.

(2) $R_{3}$ has a $\Psi_{xy}^{*}(2)$-cover.

(3) Both $R_{4}$ and $R_{5}$ have a $\Psi_{xy}(2)$-cover $\mathcal{F}$
satisfying that $y\notin V(T_{1})\cup V(T_{2})$, $x\notin V(T_{3})$ and $xy$
is in the tadpole-path of $T_{4}$, where $\\{T_{1},T_{2}\\}$ and
$\\{T_{3},T_{4}\\}$ are the sets of tadpoles of $\mathcal{F}$ at $x$ and $y$,
respectively.

By convention, for any $H=H(u,v)$, the notation $H=R_{i}$ (resp., $H\sim
R_{i}$) means that $G$ is isomorphic (resp., equivalent) to $H$ and
$\\{u,v\\}$ corresponds to $\\{x,y\\}$; while $H=R_{i}(x,y)$ (resp., $H\sim
R_{i}(x,y)$) means that $G$ is isomorphic (resp., equivalent) to $H$ and
$(u,v)$ corresponds to $(x,y)$.

######  Lemma 3.5.

Let $H=H(x,y)$ and $G=\mathcal{P}(H\cup yz,xz)$ such that $xy\in E(H)$ and
$xyzx$ is an unbalanced triangle. If $H\sim R_{i}(x,y)$ for some
$i\in\\{2,4,5\\}$ or $H$ has a $\Psi_{xy}^{*}(2)$-cover, then $G$ has a
$\Psi_{xz}^{*}(2)$-cover.

###### Proof.

With possible switching, we assume that $\sigma(xy)=1$. If
$H=\\{R_{2}(x,y),R_{4}(x,y),R_{5}(x,y)\\}$, then $G$ is a small signed graph
and thus it is easy to find a $\Psi_{xz}^{*}(2)$-cover of $G$. Now we assume
that $H$ has a $\Psi_{xy}^{*}(2)$-cover $\mathcal{F}_{H}$. By the definition
of $\Psi_{xy}^{*}(2)$-cover, let

$\mathcal{F}_{H}=\mathcal{C}_{0}\cup\\{P_{1},P_{2}\\}\cup\\{Q_{1},Q_{2}\\}\cup\\{T_{x1},xy\cup
T_{y2}\\}\cup\\{T_{y1},yx\cup T_{x2}\\},$

where $\mathcal{C}_{0}$ is a family of signed circuits, $P_{1},P_{2}$ (resp.,
$Q_{1},Q_{2}$) are two positive (resp., negative) $xy$-paths, $T_{u1},T_{u2}$
are the two tadpoles at $u$ not containing the vertex in
$\\{x,y\\}\setminus\\{u\\}$ for each $u\in\\{x,y\\}$.

Let $e_{0}=xy$, $e_{1}=xz$ and $e_{2}=zy$. Since $xyzx$ is unbalanced and
$\sigma(e_{0})=1$, WLOG, assume that $\sigma(e_{1})=-1$ and $\sigma(e_{2})=1$.
From $G$ and $\mathcal{F}\setminus\mathcal{C}_{0}$, we construct an auxiliary
signed graph $G^{\prime}$ shown in Fig. 4. Observe that the family

$\displaystyle\mathcal{F}_{G^{\prime}}=$ $\displaystyle\\{e_{3}\cup
e_{2},e_{3}\cup e_{2}\\}\cup\\{e_{1},e_{1}\\}\cup\\{e_{5},e_{1}\cup e_{2}\cup
e_{6}\\}\cup\\{e_{2}\cup e_{6},e_{1}\cup e_{0}\cup e_{4}\\}$
$\displaystyle\cup\\{e_{1}\cup e_{2}\cup e_{0}\cup e_{5},e_{1}\cup e_{2}\cup
e_{4}\\}$

covers $\\{e_{1},e_{2}\\}$ $6$ times and
$E(G^{\prime})\setminus\\{e_{1},e_{2}\\}$ twice. Let $\mathcal{F}_{G}$ be the
family obtained from $\mathcal{F}_{G^{\prime}}$ by replacing two $e_{3}$s with
$P_{1},P_{2}$, two $e_{4}$s with $Q_{1},Q_{2}$, two $e_{5}$s with
$T_{x1},T_{x2}$, two $e_{6}$s with $T_{y1},T_{y2}$. One can easily check that
$\mathcal{F}_{G}\cup\mathcal{C}_{0}$ is a $\Psi_{xz}^{*}(2)$-cover of $G$. ∎

$x$$z$$y$$e_{1}$$e_{2}$$e_{0}$$e_{3}$$e_{4}$$e_{5}$$e_{6}$ Figure 4: An
auxiliary signed graph $G^{\prime}$.

By Observation 3.4, each member in $\\{R_{2},R_{4},R_{5}\\}$ has a
$\Psi_{xy}(2)$-cover in which at least one tadpole at $y$ doesn’t contain $x$.
By this fact and a similar method of the proof of Lemma 3.5, we obtain the
following lemma.

######  Lemma 3.6.

Let $H_{i}=H_{i}(x,y_{i})$ for each $i\in[1,2]$, $G=\mathcal{P}(H_{1}\cup
y_{1}z,H_{2}\cup y_{2}z)$ and $G^{\prime}$ be the signed graph obtained from
$G$ by adding a new negative loop at $x$.

(1) If $H_{1}=R_{0}$, and either $H_{2}\sim R_{j}(x,y)$ for some
$j\in\\{0,2,4,5\\}$ or $H_{2}$ has a $\Psi_{xy_{2}}^{*}(2)$-cover, then both
$G$ and $G^{\prime}$ have a signed circuit $6$-cover. Moreover, $G$ has a
$\Psi_{xz}(t)$-cover for each $t\in[0,3]$.

(2) If either $H_{i}\sim R_{j}(x,y)$ for some $j\in\\{2,4,5\\}$ or $H_{i}$ has
a $\Psi_{xy_{i}}^{*}(2)$-cover for each $i\in[1,2]$, then both $G$ and
$G^{\prime}$ have a signed circuit $6$-cover. Moreover, $G$ has a
$\Psi_{xz}(2)$-cover in which no tadpole at $z$ contains $x$.

##  4 Proof of Theorem 1.2

In this section, we will complete the proof of Theorem 1.2 by contradiction.

Let $G$ be a counterexample to Theorem 1.2 with minimum $|E(G)|$. Then $G$ is
unbalanced since every coverable graph has a circuit $6$-cover (see [12]). By
the minimality, $G$ contains no $2$-vertices and can’t be decomposed into two
coverable signed subgraphs. The latter implies that $G$ is connected and
contains no positive loops.

###  4.1 Properties of the smallest counterexample $G$

In this subsection, we will present some properties of $G$.

For convenience, a two-terminal signed graph $H=H(x,y)$ is said to be a piece
of $G$ at $\\{x,y\\}$ if there is another two-terminal signed graph
$H^{\prime}=H^{\prime}(x,y)$ such that $G=\mathcal{P}(H,H^{\prime})=H\cup
H^{\prime}$.

######  Claim 4.1.

The following statements hold.

  * (1)

No two negative loops share a common vertex.

  * (2)

$G$ is $2$-connected.

  * (3)

Every balanced piece of $G$ is a positive or negative $K_{2}$.

  * (4)

If $R_{1}$ is a piece of $G$ at $\\{x,y\\}$, then $d_{G}(x)\geq 4$ and
$d_{G}(y)\geq 4$.

  * (5)

$G$ contains no balanced subgraph $H=K_{4}-y_{1}y_{2}$, where $K_{4}$ is the
complete graph on vertices $x,y_{1},y_{2},y_{3}$ and $x$ is a $3$-vertex of
$G$.

  * (6)

$G-L(G)$ contains no adjacent $2$-vertices.

###### Proof.

(1) Suppose to the contrary that $e_{1},e_{2}$ are two negative loops at a
vertex. Since $C_{0}=e_{1}\cup e_{2}$ is a short barbell,
$G-\\{e_{1},e_{2}\\}$ is not coverable, so $G-e_{1}$ is coverable. By the
minimality of $G$, $G-e_{1}$ has a signed circuit $6$-cover $\mathcal{F}$.
Pick three signed circuits $C_{1},C_{2},C_{3}$ from $\mathcal{F}$ containing
$e_{2}$. Then the family

$(\mathcal{F}\setminus\\{C_{1},C_{2},C_{3}\\})\cup\\{C_{1}\bigtriangleup
C_{0},C_{2}\bigtriangleup C_{0},C_{2}\bigtriangleup C_{0}\\}\cup 3\\{C_{0}\\}$

is a signed circuit $6$-cover of $G$, a contradiction. This proves (1).

(2) Suppose to the contrary that there are two subgraphs $H_{1},H_{2}$ in $G$
such that $G=H_{1}\cup H_{2}$ and $V(H_{1})\cap V(H_{2})=\\{x\\}$. Since the
minimum degree of $G$ is at least three, $|E(H_{i})|\geq 2$ for each
$i\in[1,2]$. Note that if $H_{i}$ is balanced, then it is coverable and thus
both $H_{1}$ and $H_{2}$ are coverable, a contradiction to the minimality of
$G$. Hence neither $H_{1}$ nor $H_{2}$ is balanced. Therefore for each
$i\in[1,2]$, the signed graph obtained from $H_{i}$ by adding a new negative
loop $L_{i}$ at $x$ is also coverable and thus has a signed circuit $6$-cover
$\mathcal{F}_{i}$ by the minimality of $G$ again. Let
$\mathcal{C}_{i}=\\{C_{i1},\dots,C_{i6}\\}$ be the six signed circuits in
$\mathcal{F}_{i}$ containing $L_{i}$. Then the family

$(\mathcal{F}_{1}\setminus\mathcal{C}_{1})\cup(\mathcal{F}_{2}\setminus\mathcal{C}_{2})\cup(\cup_{j=1}^{6}\\{(C_{1j}\setminus\\{L_{1}\\})\cup(C_{2j}\setminus\\{L_{2}\\})\\})$

is a signed circuit $6$-cover of $G$, a contradiction. This proves (2).

(3) Suppose to the contrary that there are two pieces $H_{1},H_{2}$ of $G$ at
$\\{x,y\\}$ such that $G=H_{1}\cup H_{2}$, $H_{2}$ is balanced, and
$|E(H_{2})|\geq 2$. Without loss of generality, assume that $H_{2}$ has a
positive $xy$-path. Then all $xy$-paths in $H_{2}$ are positive since $H_{2}$
is balanced.

For each $i=1,2$, let $H_{i}^{\prime}$ be the graph obtained from $H_{i}$ by
adding a new positive edge $e_{i}$ connecting $x$ and $y$. Then
$|E(H_{1}^{\prime})|<|E(G)|$ and $H_{2}^{\prime}$ is balanced. Moreover, both
$H_{1}^{\prime}$ and $H_{2}^{\prime}$ are $2$-connected and $K_{4}$-minor-
free. Obviously, $H_{2}^{\prime}$ has a balanced circuit $6$-cover, denoted by
$\mathcal{F}_{2}$.

We first show that $H_{1}^{\prime}$ is coverable. Suppose not. Since
$H_{1}^{\prime}$ is $2$-connected, by Proposition 2.2, there is an edge $e$ in
$H_{1}^{\prime}$ such that $H_{1}^{\prime}-e$ is balanced. Since $e_{1}$ is a
positive edge in $H_{1}^{\prime}$, every $xy$-path in $H_{1}-e$ is positive
and thus $G-e$ is balanced. Since $G$ is unbalanced, it is not coverable by
Proposition 2.2, a contradiction. Therefore $H_{1}^{\prime}$ is coverable.

By the minimality of $G$, $H_{1}^{\prime}$ has a signed circuit $6$-cover
$\mathcal{F}_{1}$. For each $i=1,2$, let
$\mathcal{C}_{i}=\\{C_{i1},\cdots,C_{i6}\\}$ be the six members of
$\mathcal{F}_{i}$ containing $e_{i}$. Since every member of $\mathcal{C}_{2}$
is a balanced circuit, the family

$(\mathcal{F}_{1}\setminus\mathcal{C}_{1})\cup(\mathcal{F}_{2}\setminus\mathcal{C}_{2})\cup\left(\cup_{i=j}^{6}\\{(C_{1j}\setminus\\{e_{1}\\})\cup(C_{2j}\setminus\\{e_{2}\\})\\}\right)$

is a signed circuit $6$-cover of $G$, a contradiction. This proves (3).

(4) By symmetry, we only need to show that $d_{G}(y)\geq 4$. Suppose by
contradiction that $d_{G}(y)=3$. Let $H=H(x,y)$ be a piece of $G$ such that
$G=\mathcal{P}(H,R_{1})$. As shown in Fig. 3, denote $C_{0}=xyzx$ and
$R_{1}=C_{0}\cup L_{z}$. Clearly $G-xy$ is $2$-connected and coverable. Thus
it has a signed circuit $6$-cover $\mathcal{G}$ by the minimality of $G$. For
each $C\in\mathcal{G}$, $E(C)\cap\\{L_{z},xz,yz\\}$ is either
$\\{xz,L_{z}\\}$, or $\\{yz,L_{z}\\}$, or $\\{xz,yz\\}$, or
$\\{xz,yz,L_{z}\\}$. Denote by $\mathcal{G}_{1}$ (resp., $\mathcal{G}_{2}$,
$\mathcal{G}_{3}$, $\mathcal{G}_{4}$) the set of signed circuits
$C\in\mathcal{G}$ with $E(C)\cap E(R_{1})=\\{xz,L_{z}\\}$ (resp.,
$=\\{yz,L_{z}\\}$, $=\\{xz,yz\\}$, $=\\{xz,yz,L_{z}\\}$). Since
$\mathcal{G}(xz)=\mathcal{G}(yz)=\mathcal{G}(L_{z})=6$, we have

$|\mathcal{G}_{1}|+|\mathcal{G}_{3}|+|\mathcal{G}_{4}|=|\mathcal{G}_{2}|+|\mathcal{G}_{3}|+|\mathcal{G}_{4}|=|\mathcal{G}_{1}|+|\mathcal{G}_{2}|+|\mathcal{G}_{4}|=6.$

Thus there is an integer $t\in[0,3]$ such that
$|\mathcal{G}_{1}|=|\mathcal{G}_{2}|=|\mathcal{G}_{3}|=t$ and
$|\mathcal{G}_{4}|=6-2t$. Let $\mathcal{G}_{i}=\\{C_{i1},\dots,C_{it}\\}$ for
each $i\in[1,3]$ and $\mathcal{G}_{4}=\\{C_{41},\dots,C_{4(6-2t)}\\}$. Then
the family

$\left\\{\begin{array}[]{ll}(\mathcal{G}\setminus\\{C_{41},C_{42},C_{43},C_{44}\\})\cup\\{C_{41}\bigtriangleup
zxy,C_{42}\bigtriangleup zxy,C_{43}\bigtriangleup xyz,C_{44}\bigtriangleup
xyz\\}\cup 2\\{C_{0}\\}&\mbox{if $t\in[0,1]$};\\\
(\mathcal{G}\setminus\\{C_{11},C_{31},C_{43},C_{44}\\})\cup\\{C_{11}\bigtriangleup
C_{0},C_{31}\bigtriangleup C_{0},C_{43}\bigtriangleup xyz,C_{44}\bigtriangleup
xyz\\}\cup 2\\{C_{0}\\}&\mbox{if $t=2$};\\\
(\mathcal{G}\setminus\\{C_{31},C_{32},C_{33}\\})\cup\\{C_{31}\bigtriangleup
C_{0},C_{32}\bigtriangleup C_{0},C_{33}\bigtriangleup C_{0}\\}\cup
3\\{C_{0}\\}&\mbox{if $t=3$}\end{array}\right.$

is a signed circuit $6$-cover of $G$, a contradiction. This proves (4).

(5) Suppose that such a balanced subgraph $H$ exists. Since $G$ is
$K_{4}$-minor-free, $y_{3}$ is in all $y_{1}y_{2}$-paths of $G-x$ and thus
$y_{3}$ is a cut vertex of $G-x$. Let $H_{1},H_{2}$ be two subgraphs of $G-x$
such that $G-x=H_{1}\cup H_{2}$, $V(H_{1})\cap V(H_{2})=\\{y_{3}\\}$ and
$y_{i}y_{3}\in E(H_{i})$ for each $i\in[1,2]$. Note that $d_{H_{i}}(y_{i})\geq
2$. Since $G$ is $2$-connected, by (3), either $H_{i}$ is $2$-connected and
unbalanced, or $H_{i}$ is the union of $y_{i}y_{3}$ and a negative loop at
$y_{i}$. Hence $G-x$ is coverable and thus has a signed circuit $6$-cover
$\mathcal{F}$ by the minimality of $G$. Pick six distinct signed circuits
$C_{11},C_{12},C_{13},C_{21},C_{22},C_{23}$ from $\mathcal{F}$ such that
$y_{i}y_{3}\in E(C_{ij})$ for each $i\in[1,2]$ and $j\in[1,3]$. Then the
family

$(\mathcal{F}\setminus(\cup_{i=1}^{2}\cup_{j=1}^{3}\\{C_{ij}\\}))\cup(\cup_{i=1}^{2}\cup_{j=1}^{3}\\{C_{ij}\bigtriangleup
xy_{i}y_{3}x\\})\cup 3\\{xy_{1}y_{3}y_{2}x\\}$

is a signed circuit $6$-cover of $G$, a contradiction. This proves (5).

(6) Suppose to the contrary that $x,y$ are two adjacent $2$-vertices of
$G-L(G)$. If $G-L_{y}$ is not coverable, then $E_{N}(G-L_{y})=\\{L_{x}\\}$ and
$G-\\{L_{x},L_{y}\\}$ is balanced. Since $G-\\{L_{x},L_{y}\\}$ is
$2$-connected, it has a balanced circuit $6$-cover $\mathcal{F}$. Pick
$C_{1},C_{2},C_{3}\in\mathcal{F}$ with $xy\in E(C_{i})$ for $i\in[1,3]$. Then
the family

$(\mathcal{F}\setminus\\{C_{1},C_{2},C_{3}\\})\cup(\cup_{i=1}^{3}\\{L_{x}\cup
xy\cup L_{y},L_{x}\cup(C_{i}-xy)\cup L_{y}\\})$

is a signed circuit $6$-cover of $G$, a contradiction. Therefore, $G-L_{y}$ is
coverable. Let $\mathcal{F}^{\prime}$ be a signed circuit $6$-cover of
$G-L_{y}$ by the minimality of $G$. Similar to the proof of (4), we can extend
$\mathcal{F}^{\prime}$ to a signed circuit $6$-cover of $G$, a contradiction.
This proves (6) and thus completes the proof of the claim. ∎

######  Claim 4.2.

Let $H$ be a $2$-connected piece of $G$. If $\epsilon(H)=1$, then $H\sim
R_{i}$ for some $i\in[0,5]$.

###### Proof.

Let $H$ and $H^{\prime}$ be two pieces of $G$ at $\\{x,y\\}$ such that
$G=\mathcal{P}(H,H^{\prime})$. Without loss of generality, assume that
$E_{N}(H)=\\{e_{0}\\}$ and the ends of $e_{0}$ are $z_{1},z_{2}$ (possibly
$z_{1}=z_{2}$). If $|V(H)|\leq 3$, it is obvious that
$H\in\\{R_{0},R_{1},R_{2}\\}$. Thus we assume $|V(H)|\geq 4$.

We first show that $H$ is outerplanar. Since $H$ is $K_{4}$-minor-free, it is
sufficient to prove that $H$ is $K_{2,3}$-minor-free. Suppose by contradiction
that $H$ has a $K_{2,3}$-minor. Then there are two distinct vertices $u,v$ and
three internally disjoint $uv$-paths $P_{1},P_{2},P_{3}$ in $H$ such that each
$|V(P_{i})|\geq 3$. For each $i\in[1,3]$, let $M_{i}$ be the component of
$H-\\{u,v\\}$ containing $V(P_{i})\setminus\\{u,v\\}$, and
$M_{i}^{\prime}=H[V(M_{i})\cup\\{u,v\\}]-uv$. Since $H$ is $K_{4}$-minor-free,
for $\\{w_{1},w_{2}\\}=\\{x,y\\}$ or $\\{z_{1},z_{2}\\}$, there are at least
two members of $\\{M_{1},M_{2},M_{3}\\}$ containing neither $w_{1}$ nor
$w_{2}$. Therefore there is an index in $[1,3]$, say $3$, such that
$\\{x,y,z_{1},z_{2}\\}\cap V(M_{3})=\emptyset$. This implies that
$M_{3}^{\prime}$ is an all-positive piece of $G$ at $\\{u,v\\}$. By Claim
4.1-(3), $M_{3}^{\prime}$ is a positive or negative $K_{2}$, a contradiction
to $|E(M_{3}^{\prime})|\geq|E(P_{3})|=|V(P_{i})\setminus\\{u,v\\}|+1\geq 2$.
This proves that $H$ is outerplanar.

Let $C$ be an outer facial circuit of $H$. Similar to the above argument, we
can show that $xy\in E(C)$ and either $e_{0}$ is a negative loop not at
$\\{x,y\\}$ or there is an edge $e_{1}\in E(H)$ such that $e_{0}\cup e_{1}$ is
an unbalanced $2$-circuit and $E(C)\cap\\{e_{0},e_{1}\\}\neq\emptyset$.
Without loss of generality, assume that $z_{1},x,y,z_{2}$ appear on $C$ in the
cyclic order. Let

$P_{1}=z_{1}Cx=u_{0}u_{1}\cdots u_{p}\mbox{ and }P_{2}=yCz_{2}=v_{q}\cdots
v_{1}v_{0}$

such that $C=z_{1}z_{2}\cup P_{1}\cup xy\cup P_{2}$, where $u_{0}=z_{1}$,
$u_{p}=x$, $v_{q}=y$ and $v_{0}=z_{2}$. Because $H$ is outerplanar and the
minimum degree of $G$ is at least $3$, $E(H)\setminus E(C)\subseteq
E_{H}[V(P_{1}),V(P_{2})]$ by Claim 4.1-(3).

If $u_{0}=v_{0}$, then $e_{0}$ is a negative loop not at $\\{x,y\\}$. Thus
$u_{0}u_{1}v_{1}u_{0}\cup e_{0}=R_{1}$. Since $|V(C)|=|V(H)|\geq 4$, either
$d_{G}(u_{1})\leq 3$ or $d_{G}(v_{1})\leq 3$, contradicting Claim 4.1-(4).
Hence $u_{0}\neq v_{0}$.

Note that no two indices $i\in[0,p]$ and $j\in[0,q-2]$ with
$\\{v_{j},v_{j+1},v_{j+2}\\}\subseteq N_{H}(u_{i})$. Otherwise
$H[\\{u_{i},v_{j},v_{j+1},v_{j+2}\\}]-e_{0}=K_{4}-v_{j}v_{j+2}$ and
$d_{G}(v_{j+1})=d_{H}(v_{j+1})=3$, where $K_{4}$ is the complete graph on
$\\{u_{i},v_{j},v_{j+1},v_{j+2}\\}$, contradicting Claim 4.1-(5). By the
symmetry of $P_{1}$ and $P_{2}$, it follows that $p\geq 1$, $q\geq 1$ and

$\left\\{\begin{array}[]{ll}d_{H}(w)=3&\mbox{if $w\in\\{u_{p},v_{q}\\}$;}\\\
d_{H}(w)\leq 4&\mbox{if $w\in
V(H)\setminus\\{u_{p},v_{q}\\}$.}\end{array}\right.$ (1)

Let $H^{*}=H[\\{u_{0},u_{1},v_{0},v_{1}\\}]$. According to $d_{H}(u_{0})$ and
$d_{H}(v_{0})$, we distinguish the following two cases.

Case 1. $d_{H}(u_{0})=4$ or $d_{H}(v_{0})=4$.

By the symmetry of $P_{1}$ and $P_{2}$, assume that $d_{H}(v_{0})=4$. Then
$d_{H}(u_{0})=3$, $u_{1}v_{0}\in E(H)$ and $H[\\{u_{0},u_{1},v_{0}\\}]=R_{1}$.
If $p=1$, then $H=H^{*}=R_{4}$ by Eq. (1). Now we assume $p\geq 2$.

If $d_{H}(u_{1})=3$, then $u_{2}u_{1}\cup H[\\{u_{0},u_{1},v_{0}\\}]$ is a
piece of $G$ at $\\{u_{2},v_{0}\\}$ and has a $\Psi_{v_{0}u_{2}}(t)$-cover for
each $t\in[0,3]$ in which no tadpole at $v_{0}$ contains $u_{2}$. By Lemma
3.3-(1) and the minimality of $G$, $G$ has a signed circuit $6$-cover, a
contradiction.

If $d_{H}(u_{1})=4$, then $H^{*}=R_{4}$. Let $C_{0}=u_{1}v_{0}v_{1}u_{1}$ and
$G^{\prime}=G-\\{u_{1}v_{0},u_{1}v_{1}\\}$. Clearly, $G^{\prime}$ has a signed
circuit $6$-cover $\mathcal{F}^{\prime}$ by the minimality of $G$. Note that
$d_{G^{\prime}}(u_{1})=2$ and $d_{G^{\prime}}(u_{0})=d_{G^{\prime}}(v_{0})=3$.
By the structure of $G^{\prime}$, there are $3$ signed circuits
$C_{1},C_{2},C_{3}$ in $\mathcal{F}^{\prime}$ such that each of $C_{1},C_{2}$
contains the tadpole $e_{0}\cup e_{1}\cup v_{0}v_{1}$ but not the vertex
$u_{1}$, and $C_{3}$ contains the path $u_{2}u_{1}u_{0}\cup e_{1}\cup
v_{0}v_{1}$. Hence the family

$(\mathcal{F}^{\prime}\setminus\\{C_{1},C_{2},C_{3}\\})\cup\\{C_{1}\bigtriangleup
C_{0},C_{2}\bigtriangleup C_{0},C_{3}\bigtriangleup C_{0}\\}\cup 3\\{C_{0}\\}$

is a signed circuit $6$-cover of $G$, a contradiction.

Case 2. $d_{H}(u_{0})=d_{H}(v_{0})=3$.

If $p=q=1$, then $H=R_{3}$. If $p=1$ and $q\geq 2$, then $d_{H}(v_{1})=3$ by
Eq. (1) and thus $q=2$ and $H=R_{5}$. By the symmetry of $P_{1}$ and $P_{2}$,
we assume that $p\geq 2$ and $q\geq 2$. Then $u_{1}v_{1}\in E(H)$.

If $d_{H}(u_{1})=d_{H}(v_{1})=3$, then $H^{*}=R_{3}$. By Lemma 3.2-(1)
$u_{2}u_{1}\cup H^{*}\cup v_{1}v_{2}$ has a $\Psi_{u_{2}v_{2}}(2)$-cover
satisfying the condition of Lemma 3.3-(2). Together with the minimality of
$G$, we can obtain a signed circuit $6$-cover of $G$, a contradiction.

Assume that either $d_{H}(u_{1})\geq 4$ or $d_{H}(v_{1})\geq 4$. WLOG assume
$d_{H}(u_{1})\geq 4$. Let $G^{\prime}=G-\\{u_{1}v_{1},u_{1}v_{2}\\}$. With a
similar argument in the case when $d_{H}(u_{1})=4$ in Case 1, one can find a
signed circuit $6$-cover of $G$, a contradiction. This completes the proof of
the claim. ∎

For two distinct $x,y\in V(G)$, let $t_{G}(x,y)$ denote the maximum number of
pieces $H_{1},\dots,H_{t}$ of $G$ at $\\{x,y\\}$ such that
$G=\mathcal{P}(H_{1},\dots,H_{t})$.

######  Claim 4.3.

$t_{G}(x,y)\leq 3$ for any two distinct vertices $x,y\in V(G)$.

###### Proof.

Suppose to the contrary that there are two distinct vertices $x,y$ such that
$t=t_{G}(x,y)\geq 4$. Let $H_{1},\dots,H_{t}$ be $t$ pieces of $G$ at
$\\{x,y\\}$ such that $G=\mathcal{P}(H_{1},\dots,H_{t}).$ Since $G$ is
$K_{4}$-minor-free, no $H_{i}$ is $2$-connected by the maximality of $t$.

Without loss of generality, assume
$\epsilon(H_{1})\leq\epsilon(H_{2})\leq\cdots\leq\epsilon(H_{t})$. Then by
Claim 4.1-(3), $\epsilon(H_{1}\cup H_{2})\geq 1$ and $\epsilon(H_{3})\geq 1$.
Thus if $t\geq 5$, then $G$ can be decomposed into two coverable subgraphs
$H_{1}\cup H_{2}\cup H_{3}$ and $H_{4}\cup\cdots\cup H_{t}$, a contradiction.
Hence $t=4$.

We first consider the case when $\epsilon(H_{2})=0$. Then $H_{1}\cup
H_{2}=R_{0}$ by Claim 4.1-(3). Since $G$ cannot not be decomposed into two
coverable subgraphs, $H_{i}\cup H_{j}$ is not coverable for some
$i\in\\{1,2\\}$ and $j\in\\{3,4\\}$. WLOG, assume that $H_{1}\cup H_{3}$ is
not coverable and thus $\epsilon(H_{1}\cup H_{3})=1$. By Claim 4.2, for some
$k\in[1,5]$,

$H_{1}\cup H_{3}\sim R_{k}\mbox{ and }H_{1}\cup H_{2}\cup H_{3}\sim R_{k}\cup
e,$ (2)

where $e$ is a negative edge not in $R_{k}$ with ends $x,y$. Since $H_{1}\cup
H_{2}\cup H_{3}$ is coverable, $H_{4}$ is not coverable. Thus by Proposition
2.2, there is an edge $b$ of $H_{4}$ such that $H_{4}-b$ has a balanced
component $M$. Since $H_{4}$ is not $2$-connected, it follows from Claim
4.1-(3) that $M$ is the single vertex $x$ or $y$, say $y$. Then
$d_{H_{4}}(y)=1$. Let $b=yy^{\prime}$. Then $H=H_{1}\cup H_{2}\cup H_{3}\cup
b$ is a piece of $G$ at $\\{x,y^{\prime}\\}$. By Eq. (2), it is easy to check
that for any $t\in[0,3]$, $H$ has a $\Psi_{xy^{\prime}}(t)$-cover in which no
tadpole at $x$ contains $y^{\prime}$. By Lemma 3.3-(1) and the minimality of
$G$, $G$ has a signed circuit $6$-cover, a contradiction.

Now we consider the case when $\epsilon(H_{2})\geq 1$. Since
$\epsilon(H_{2})\geq 1$, for any $\\{i,j\\}\subseteq[2,4]$,
$\epsilon(H_{i}\cup H_{j})\geq\epsilon(H_{i})+\epsilon(H_{j})\geq 2$. Thus
$H_{i}\cup H_{j}$ is coverable. This implies that for each $j\in[2,4]$,
$H_{1}\cup H_{j}$ is not coverable and thus by Claim 4.2, $H_{1}\cup H_{j}\sim
R_{k_{j}}$ for some $k_{j}\in[1,5]$. With some switchings, assume that $H_{1}$
is the positive edge $xy$. Thus for each $j\in[2,4]$,

$H_{j}=R_{k_{j}}-xy.$

One can check directly that $G$ has a signed circuit $6$-cover, a
contradiction. This proves the claim. ∎

######  Claim 4.4.

$H-L(H)$ is unbalanced for every $2$-connected piece $H$ of $G$.

###### Proof.

Prove by contradiction. Let $x,y$ be two distinct vertices and $H$ be a piece
of $G$ at $\\{x,y\\}$ such that

  * (i)

$H-L(H)$ is balanced;

  * (ii)

subject to (i), $|E(H)|$ is as small as possible.

Without loss of generality, assume that $H-L(H)$ is all-positive. Then
$E_{N}(H)=L(H)$. By (ii), no member of $L(H)$ has end $x$ or $y$. Denote by
$H^{\prime}$ another piece of $G$ at $\\{x,y\\}$ such that
$G=\mathcal{P}(H,H^{\prime})$.

We first show $H=R_{1}$. It is sufficient to prove that $H-L(H)$ is a
$3$-circuit. Suppose not. Then either $H-L(H)=xz_{1}yz_{2}x$ is a $4$-circuit
by Claim 4.1-(6) or $\overline{H-L(H)}$ has minimum degree at least $3$. In
the former case , $H=xz_{1}yz_{2}x\cup\\{L_{z_{1}},L_{z_{2}}\\}$ and hence any
signed circuit $6$-cover of $H^{\prime}\cup\\{L_{x},L_{y}\\}$ can be extended
to a signed circuit $6$-cover of $G$, where $L_{u}$ is a new negative loop at
$u$ for $u\in\\{x,y\\}$, a contradiction. In the latter case, since $G=H\cup
H^{\prime}$ is $2$-connected and $K_{4}$-minor-free, $\overline{G-L(H)}$
contains a $2$-circuit, which corresponds to a $2$-connected piece of $H$ (and
thus $G$) satisfying (i), a contradiction to (ii). Thus $H=R_{1}$.

Next we show that $H^{\prime}$ has a cut-edge. Otherwise $H^{\prime}$ is
$2$-edge-connected. Since $G=\mathcal{P}(H,H^{\prime})$ and $H$ is
$2$-connected, $t_{H^{\prime}}(x,y)=t_{G}(x,y)-t_{H}(x,y)\leq 1$ by Claim 4.3.
Thus $H^{\prime}$ contains cut vertices separating $x$ from $y$. This implies
that there are $s$ $(\geq 2)$ $2$-connected subgraphs or negative loops
$B_{1},\dots,B_{s}$ such that $H^{\prime}=\mathcal{S}(B_{1},\dots,B_{s})$ with
$x\in V(B_{1})$ and $y\in V(B_{s})$. By Claim 4.1-(3), $\epsilon(B_{i})\geq 1$
for each $i\in[1,s]$. Moreover we have $s=2$ and
$\epsilon(B_{1})=\epsilon(B_{2})=1$ since $G$ can’t be decomposed into two
coverable subgraphs. By Claim 4.2, $B_{1}\sim R_{j_{1}}$ and $B_{2}\sim
R_{j_{2}}$ for some $j_{1},j_{2}\in[0,5]$. Since $H=R_{1}$, one can find a
signed circuit $6$-cover of $G$, a contradiction. Thus $H^{\prime}$ has a cut-
edge.

By the above two claims, let $H=C_{0}\cup L_{z}$ where $C_{0}=xzyx$, let $uv$
be a cut-edge of $H^{\prime}$, and let $M_{1},M_{2}$ be two components of
$H^{\prime}-uv$ with $x,u\in V(M_{1})$ and $y,v\in V(M_{2})$.

Let $G^{\prime}=G-xy$. Then $G^{\prime}$ is $2$-connected and coverable. By
the minimality of $G$, $G^{\prime}$ has a signed circuit $6$-cover. Choose a
signed circuit $6$-cover $\mathcal{F}^{\prime}$ of $G^{\prime}$ such that the
number of balanced circuits and short barbells in $\mathcal{F}^{\prime}$ is as
large as possible.

To complete the proof, we will construct a signed circuit $6$-cover
$\mathcal{F}$ of $G$ from $\mathcal{F}^{\prime}$.

With a similar argument of the proof of Claim 4.1-(4), one can show that there
is an integer $t\in[0,3]$ and four families
$\mathcal{F}_{i}=\\{C_{i1},\dots,C_{it_{i}}\\}$, $i\in[1,4]$, in
$\mathcal{F}^{\prime}$ such that $t_{1}=t_{2}=t_{3}=t$, $t_{4}=6-2t$ and for
every $C\in\mathcal{F}_{1}$ (resp., $\mathcal{F}_{2}$, $\mathcal{F}_{3}$,
$\mathcal{F}_{4}$), $E(C)\cap E(H)=\\{L_{z},zx\\}$ (resp., $=\\{L_{z},yz\\}$,
$=\\{zx,yz\\}$, $=\\{L_{z},zx,yz\\}$).

If $t\in[0,1]$, let

$\mathcal{F}=(\mathcal{F}^{\prime}\setminus\\{C_{41},C_{42},C_{43},C_{44}\\})\cup\\{C_{41}\bigtriangleup
zxy,C_{42}\bigtriangleup zxy,C_{43}\bigtriangleup xyz,C_{44}\bigtriangleup
xyz\\}\cup 2\\{C_{0}\\}.$

If $t=3$, let
$\mathcal{F}=(\mathcal{F}^{\prime}\setminus\\{C_{31},C_{32},C_{33}\\})\cup\\{C_{31}\bigtriangleup
C_{0},C_{32}\bigtriangleup C_{0},C_{33}\bigtriangleup C_{0}\\}\cup
3\\{C_{0}\\}.$

If $t=2$ and either $y\notin V(C_{11})\cap V(C_{12})$ or $x\notin
V(C_{21})\cap V(C_{22})$, say $y\notin V(C_{11})$, let

$\mathcal{F}=(\mathcal{F}^{\prime}\setminus\\{C_{11},C_{31},C_{41},C_{42}\\})\cup\\{C_{11}\bigtriangleup
C_{0},C_{31}\bigtriangleup C_{0},C_{41}\bigtriangleup xyz,C_{42}\bigtriangleup
xyz\\}\cup 2\\{C_{0}\\}.$

In each of the above case, we obtain a signed circuit $6$-cover of $G$, a
contradiction.

Finally we consider the case that $t=2$, $y\in V(C_{11})\cap V(C_{12})$ and
$x\in V(C_{21})\cap V(C_{22})$.

Then $uv\in\cap_{j=1}^{2}\left(E(C_{1j})\cap E(C_{2j})\cap E(C_{4j})\right)$
but $uv\notin E(C_{31})\cup E(C_{32})$. For each $j\in[1,2]$, denote by
$P_{1j}$ (resp., $T_{2j}$, $P_{4j}^{1}$), the segment of $C_{1j}$ (resp.,
$C_{2j}$, $C_{4j}$) in $M_{1}$, and by $T_{1j}$ (resp., $P_{2j}$,
$P_{4j}^{2}$) the segment of $C_{1j}$ (resp., $C_{2j}$, $C_{4j}$) in $M_{2}$.
Thus

$C_{1j}=L_{z}\cup zx\cup P_{1j}\cup uv\cup T_{1j},\ C_{2j}=L_{z}\cup zy\cup
P_{2j}\cup vu\cup T_{2j},\ C_{4j}=L_{z}\cup zx\cup P_{4j}^{1}\cup uv\cup
P_{4j}^{2}\cup yz.$

Clearly $P_{1j}$ and $P_{4j}^{1}$ are $xu$-paths, $P_{2j}$ and $P_{4j}^{2}$
are $vy$-paths, and $T_{1j}$ (resp., $T_{2j}$) is a tadpole at $v$ (resp.,
$u$).

Since $uv\notin E(C_{31})$ and $z$ is a cut vertex of $G^{\prime}-uv$,
$C_{31}$ is a barbell and $z$ is in the barbell-path of $C_{31}$. Hence there
are two barbells, denoted by $C_{31}^{1},C_{31}^{2}$, in $C_{31}\cup L_{z}$
such that $\\{C_{31}^{1},C_{31}^{2}\\}$ covers $C_{31}$ once and $L_{z}$
twice.

If
$\sigma(P_{1j_{1}})\sigma(P_{2j_{2}})\neq\sigma(P_{41}^{1})\sigma(P_{41}^{2})$
for some $j_{1},j_{2}\in[1,2]$, then $C_{1}=zx\cup P_{1j_{1}}\cup uv\cup
P_{2j_{2}}\cup yz$ is a balanced circuit. Let

$\mathcal{F}^{\prime\prime}=(\mathcal{F}^{\prime}\setminus\\{C_{1j_{1}},C_{2j_{2}},C_{31}\\})\cup\\{C_{1},C_{31}^{1},C_{31}^{2},T_{1j_{1}}\cup
uv\cup T_{2j_{2}}\\}.$

If $\sigma(P_{1j_{1}})\sigma(P_{2j_{2}})=\sigma(P_{41}^{1})\sigma(P_{41}^{2})$
for any $j_{1},j_{2}\in[1,2]$, then both $C_{2}=zx\cup P_{11}\cup uv\cup
P_{21}\cup yz$ and $C_{3}=zx\cup P_{12}\cup uv\cup P_{22}\cup yz$ are
unbalanced circuits. Let

$\mathcal{F}^{\prime\prime}=(\mathcal{F}^{\prime}\setminus\\{C_{11},C_{12},C_{21},C_{22},C_{31}\\})\cup\\{C_{2}\cup
L_{z},C_{3}\cup L_{z},C_{31}^{1},C_{31}^{2},T_{11}\cup vu\cup
T_{21},T_{12}\cup vu\cup T_{22}\\}.$

In both cases, $\mathcal{F}^{\prime\prime}$ is a signed circuit $6$-cover of
$G^{\prime}$ with a larger number of balanced circuits and short barbells than
$\mathcal{F}^{\prime}$, a contradiction to the choice of
$\mathcal{F}^{\prime}$. This completes the proof of the claim. ∎

######  Claim 4.5.

Every balanced $3$-circuit is in a piece $H$ of $G$ with $H\sim R_{i}$ for
some $i\in\\{2,4,5\\}$.

###### Proof.

Let $C=xyzx$ be a balanced triangle. For any $\\{u,v\\}\subseteq V(C)$,

$V_{uv}=\\{w\in V(G)\setminus V(C):\mbox{ there is a $uv$-path containing $w$
but not the third vertex $V(C)$ in $G$}\\}.$

Since $G$ is $2$-connected and $K_{4}$-minor-free,
$\\{V_{xy},V_{xz},V_{yz}\\}$ is a partition of $V(G)\setminus V(C)$. Let
$G_{uv}=G[V_{uv}\cup\\{u,v\\}]$, where every loop at $V(C)$ belongs to exactly
one of $\mathcal{G}=\\{G_{xy},G_{xz},G_{yz}\\}$. Then $G_{uv}$ is a piece of
$G$ at $\\{u,v\\}$ and

$G=\mathcal{P}(\mathcal{S}(G_{xz},G_{zy}),G_{xy})=G_{xz}\cup G_{zy}\cup
G_{xy}.$

Without loss of generality, assume that
$\epsilon(G_{xy})\geq\epsilon(G_{yz})\geq\epsilon(G_{xz})$. Note that, by the
definition and Claim 4.1-(3), every $G_{uv}\in\mathcal{G}$ is a positive edge
if $\epsilon(G_{uv})=0$ and is $2$-connected if $\epsilon(G_{uv})\geq 1$.

If $\epsilon(G_{xz})\geq 1$, then $\epsilon(G_{xy})\
=\epsilon(G_{yz})=\epsilon(G_{xz})=1$; otherwise $G$ can be decomposed into
two coverable subgraphs $G_{xy}$ and $G_{xz}\cup G_{yz}$, a contradiction. By
Claim 4.2, every $G_{uv}\in\mathcal{G}$ is equivalent to $R_{i}$ for some
$i\in[0,5]$. One can check easily that $G$ has a signed circuit $6$-cover, a
contradiction. Therefore $\epsilon(G_{xz})=0$. By Claim 4.1-(3), $G_{xz}=xz$.

Note that $G_{yz}\not=yz$ otherwise $z$ is a $2$-vertex of $G$. Thus
$\epsilon(G_{yz})\geq 1$.

If $\epsilon(G_{yz})\geq 2$, then $\epsilon(G_{xy})\geq\epsilon(G_{yz})\geq
2$. This implies that both $G_{xy}$ and $G_{yz}$ are coverable. By the
minimality of $G$, let $\mathcal{F}_{1}$ and $\mathcal{F}_{2}$ be two signed
circuit $6$-covers of $G_{xy}$ and $G_{yz}$, respectively. For each
$i\in[1,2]$, pick three members $C_{i1}$, $C_{i2}$, $C_{i3}$ from
$\mathcal{F}_{i}$ such that $xy\in E(C_{1j})$ and $yz\in E(C_{2j})$ for
$j\in[1,3]$. Then

$\cup_{i=1}^{2}\left((\mathcal{F}_{i}\setminus\\{C_{i1},C_{i2},C_{i3}\\})\cup\\{C_{i1}\bigtriangleup
C,C_{i2}\bigtriangleup C,C_{i3}\bigtriangleup C\\}\right)$

is a signed circuit $6$-cover of $G$, a contradiction. Therefore
$\epsilon(G_{yz})=1$.

Since $C$ is balanced, $\epsilon(G_{yz}\cup C)=\epsilon(G_{yz})=1$. By Claims
4.2 and 4.4, each $H\in\\{G_{yz},G_{yz}\cup C\\}$ is equivalent to an $R_{i}$
for some $i\in\\{0,2,3,4,5\\}$. Therefore $G_{yz}\cup C\sim R_{i}$ for some
$i\in\\{2,4,5\\}$. This completes the proof of the claim. ∎

######  Claim 4.6.

Let $B_{i}=B_{i}(x_{i-1},x_{i})$ for $i\in[1,h]$ and
$H=\mathcal{S}(B_{1},\dots,B_{h})$ be a piece of $G$ at $\\{x_{0},x_{h}\\}$
such that $h=|\mathcal{B}(H)|\geq 2$, $\epsilon(G-E(H))\geq 1$, and every
$B_{i}\in\mathcal{B}_{2}(H)$ has a $\Psi_{x_{i-1}x_{i}}(2)$-cover.

  * (1)

If $\mathcal{B}_{0}(H)=\emptyset$, then either $H\sim D_{2}(x,y)$ in Fig. 2,
or $h=2$ and $\mathcal{B}_{2}(H)\in\\{\\{B_{1}\\},\\{B_{2}\\}\\}$.
Furthermore, when $\mathcal{B}_{2}(H)=\\{B_{1}\\}$, the following statements
hold.

    * (1a)

Every $\Psi_{x_{0}x_{1}}(2)$-cover of $B_{1}$ has a tadpole at $x_{1}$
containing $x_{0}$;

    * (1b)

$B_{1}$ has no $\Psi_{x_{0}x_{1}}(t)$-cover for some $t\in\\{0,1,3\\}$;

    * (1c)

If $B_{1}$ has a $\Psi_{x_{0}x_{1}}^{*}(2)$-cover and $e=x_{0}x_{h}\in
E(G-E(H))$, then either $H\cup e$ has a $\Psi^{*}_{x_{0}x_{2}}(2)$-cover, or
$H\cup e$ is equivalent to one of $R_{2}(y,x)$, $R_{4}(x,y)$, and
$R_{5}(x,y)$.

  * (2)

If $h\geq 3$, $\mathcal{B}_{2}(H)=\\{B_{k}\\}$ and
$\mathcal{B}_{0}(H)=\\{B_{k+1}\\}$ for some $k\in[1,h-2]$, then $B_{k}$ has no
$\Psi^{*}_{x_{k-1}x_{k}}(2)$-cover and $B_{k}$ is not equivalent to $R_{i}$
for each $i\in\\{2,4,5\\}$.

###### Proof.

Let $H^{\prime}$ be a piece of $G$ at $\\{x_{0},x_{2}\\}$ such that
$G=\mathcal{P}(H,H^{\prime})$. Then $E(H^{\prime})=E(G-E(H))$.

(1) Assume $H\not\sim D_{2}(x,y)$. Since $\mathcal{B}_{0}(H)=\emptyset$ and
$\epsilon(H^{\prime})\geq 1$, if $H$ has a $\Psi_{x_{0}x_{h}}(2)$-cover in
which no tadpole at $x_{0}$ (resp., $x_{h}$) contains $x_{h}$ (resp.,
$x_{0}$), then $|E(H)|\geq 5$ and so it follows from the minimality of $G$ and
Lemma 3.3-(2) that $G$ has a signed circuit $6$-cover, a contradiction. Hence
$H$ has no such $\Psi_{x_{0}x_{h}}(2)$-cover. By Lemma 3.2, $h=2$ and
$\mathcal{B}_{2}(H)\in\\{\\{B_{1}\\},\\{B_{2}\\}\\}$.

Assume $\mathcal{B}_{1}(H_{1})=\\{B_{1}\\}$. Clearly (1a) follows from Lemma
3.2 and (1b) follows from Lemma 3.3-(1).

We now prove (1c). Suppose to the contrary that $H\cup e$ has no
$\Psi^{*}_{x_{0}x_{2}}(2)$-cover and $H\cup e$ is not equivalent to any of
$R_{2}(y,x)$, $R_{4}(x,y)$, and $R_{5}(x,y)$. Furthermore since $x_{2}$ is a
$2$-vertex of $H\cup e$, $H\cup e\not\sim R_{i}$ for each $i\in\\{2,4,5\\}$.
Since $B_{1}$ has a $\Psi^{*}_{x_{0}x_{1}}(2)$-cover, $x_{0}x_{1}\in E(B_{1})$
by the definition. With some switchings, assume that $x_{0}x_{1}$ is positive.
By Lemma 3.5, $C=x_{0}x_{1}x_{2}x_{0}$ is a balanced $3$-circuit. Note that
$x_{1}$ is a $2$-vertex of $H^{\prime}\cup C$. By Claim 4.5, $(H^{\prime}\cup
C)(x_{0},x_{1})\sim R_{2}(y,x)$ or $R_{i}(x,y)$ for some $i\in\\{4,5\\}$, and
thus $H^{\prime}\sim R_{i}(x,y)$ for some $i\in\\{0,2,3\\}$. Since
$G=\mathcal{P}(B_{1}\cup x_{1}x_{2},H^{\prime})$, by Lemma 3.6 and Observation
3.4-(2), $G$ has a signed circuit $6$-cover, a contradiction. This proves
(1c).

(2) Suppose to the contrary that either $B_{k}$ has an
$\Psi^{*}_{x_{k-1}x_{k}}(2)$-cover or $B_{k}\sim R_{i}$ for some
$i\in\\{2,4,5\\}$. Since $B_{k+1}$ is a negative loop at $x_{k}$ ($=x_{k+1}$),
$B_{k}\cup B_{k+1}$ has a $\Psi_{x_{k-1}x_{k+1}}(2)$-cover in which no tadpole
at $x_{k+1}$ contains $x_{k-1}$. Since $k\leq h-2$, we have
$B_{h}\in\mathcal{B}_{1}(H)$. Thus $H$ has a $\Psi_{x_{0}x_{h}}(2)$-cover in
which no tadpole at $x_{0}$ (resp, $x_{h}$) contains $x_{h}$ (resp., $x_{0}$).
Since $\epsilon(H^{\prime})\geq 1$, by Lemma 3.3-(2) and the minimality of
$G$, $G$ has a signed circuit $6$-cover, a contradiction. This prove (2) and
thus completes the proof of the claim. ∎

######  Claim 4.7.

Suppose that $G=\mathcal{P}(H_{1},H_{2},H_{3})$ where each $H_{i}=H_{i}(x,y)$
and $\epsilon(H_{3})\geq 1$. If $H=H_{1}\cup H_{2}\not\sim R_{i}$ for any
$i\in\\{0,2,4,5\\}$ and contains no negative loop at $\\{x,y\\}$, then the
following statements hold.

  * (1)

If either $H_{1}\sim D_{1}(x,y)$ or $H_{2}\sim D_{1}(x,y)$, then $H$ has a
$\Psi_{xy}(t)$-cover for each $t\in[0,3]$, where $D_{1}(x,y)$ is the two-
terminal signed graph in Fig. 2.

  * (2)

If $xy\in E(H)$, then $H$ has a $\Psi_{xy}^{*}(2)$-cover.

  * (3)

If $xy\notin E(H)$ and neither $H_{1}$ nor $H_{2}$ is equivalent to
$D_{1}(x,y)$, then $H$ has a $\Psi_{xy}(2)$-cover in which no tadpole at $y$
contains $x$.

###### Proof.

Suppose that $H$ is a counterexample to the claim with minimum $|E(H)|$.
Recall that $G$ is $2$-connected and contains no positive loop. By the
definition, let $B_{i}=B_{i}(x_{i-1},x_{i})$, $i\in[1,s]$ such that

$H_{1}(x,y)=\mathcal{S}(B_{1},\dots,B_{h})=B_{1}\cup\cdots\cup B_{h},\ \
H_{2}(y,x)=\mathcal{S}(B_{h+1},\dots,B_{s})=B_{h+1}\cup\cdots\cup B_{s}$

and $s$ is maximum with this property, where $x=x_{0}=x_{s}\in V(B_{1})\cap
V(B_{s})$ and $y=x_{h}\in V(B_{h})\cap V(B_{h+1})$. Then, for any
$B\in\mathcal{B}_{2}(H_{1})\cup\mathcal{B}_{2}(H_{2})$ with terminals
$\\{u,v\\}$, $B$ is $2$-connected by the maximality of $s$, and $B-L(B)$ is
unbalanced by Claim 4.4. Furthermore, it follows from the minimality of $H$
that $B$ has either a $\Psi_{uv}(t)$-cover for each $t\in[0,3]$, or a
$\Psi_{uv}^{*}(2)$-cover, or a $\Psi_{uv}(2)$-cover in which no tadpole at $v$
contains $u$, unless $B\sim R_{i}$ for some $i\in\\{0,2,4,5\\}$. By this fact
and Observation 3.4, $B$ has a $\Psi_{u,v}(2)$-cover.

We will find a desired $\Psi_{xy}(2)$-cover of $H$, contradicting that $H$ is
a counterexample to the claim. To do this, when $H_{i}$, $i\in[1,2]$, is not a
single edge (that is, $|\mathcal{B}_{0}(H_{i})|+|\mathcal{B}_{2}(H_{i})|\geq
1$), we apply Lemma 3.1 to construct a signed subgraph $6$-cover
$\mathcal{F}_{i}^{*}$ of $H_{i}$ as follows:

$\mathcal{F}_{i}^{*}=\mathcal{F}_{i0}\cup
2\mathcal{B}_{0}(H_{i})\cup\\{P_{i1},P_{i2},P_{i3},P_{i4}\\}\cup\\{T_{i1},T_{i2},T_{i3},T_{i4}\\},$

where

  * $\rhd$

$\mathcal{F}_{i0}$ is a subfamily of signed circuits of $H_{i}$;

  * $\rhd$

$P_{i1}$ and $P_{i2}$ (resp., $P_{i3}$ and $P_{i4}$) are two positive (resp.,
negative) $xy$-paths of $H_{i}$ if $|\mathcal{B}_{2}(H_{i})|\geq 1$, and
otherwise $P_{i1}=P_{i2}=P_{i3}=P_{i4}=H_{i}-\mathcal{B}_{0}(H_{i})$;

  * $\rhd$

$T_{i1},T_{i2}$ (resp., $T_{i3},T_{i4}$) are two tadpoles of $H_{i}$ at $x$
(resp., $y$) such that the unbalanced circuit in $T_{i(2i-1)}$ (resp.,
$T_{i(2i)}$, $T_{i(5-2i)}$, $T_{i(6-2i)}$) is in the part in
$\mathcal{B}_{0}(H_{i})\cup\mathcal{B}_{2}(H_{i})$ with minimum (resp.,
minimum, maximum, maximum) subscript.

Note that $P_{11}\cup P_{21}$ is a circuit and every part in
$\mathcal{B}_{0}(H_{1})\cup\mathcal{B}_{0}(H_{2})$ is a negative loop. When
$\mathcal{B}_{0}(H_{1})\cup\mathcal{B}_{0}(H_{2})\neq\emptyset$, the signed
graph $P_{11}\cup P_{21}\cup\mathcal{B}_{0}(H_{1})\cup\mathcal{B}_{0}(H_{2})$
has a family

$\mathcal{C}_{0}\cup\\{T_{1}^{\prime},T_{2}^{\prime}\\}$

which covers $P_{11}\cup P_{21}$ once and
$\mathcal{B}_{0}(H_{1})\cup\mathcal{B}_{0}(H_{2})$ twice, where
$\mathcal{C}_{0}$ is a set of barbells and $T_{1}^{\prime},T_{2}^{\prime}$ are
two tadpoles at $x$.

(1) WLOG, assume that $H_{2}=D_{1}(x,y)$. Then $h\geq 2$ since $H\not\sim
R_{2}$. Let $t\in[0,3]$.

If $\mathcal{B}_{2}(H_{1})=\emptyset$, then $H_{1}=xx_{1}y\cup L_{x_{1}}$ by
Claim 4.1-(6), and thus it is easy to check that $H=H_{1}\cup H_{2}$ has a
$\Psi_{xy}(t)$-cover.

If $h=2$ and $B_{2}\in\mathcal{B}_{1}(H_{1})$, then
$\mathcal{B}_{0}(H_{1})=\emptyset$ and $\mathcal{B}_{2}(H_{1})=\\{B_{1}\\}$.
By (1a) and (1b) of Claim 4.6, $B_{1}$ has a $\Psi_{x_{0}x_{1}}^{*}(2)$-cover
and so $H$ has a $\Psi_{xy}(t)$-cover by Lemma 3.6-(1).

Next assume that either $h\geq 3$ and $\mathcal{B}_{2}(H_{1})\neq\emptyset$,
or $h=2$ and $B_{2}\in\mathcal{B}_{2}(H_{1})$. Then $x\notin V(T_{13})\cup
V(T_{14})$. We construct a family $\mathcal{F}^{*}$ as follows.

$\displaystyle\mathcal{F}^{*}$
$\displaystyle=\mathcal{F}_{10}\cup\mathcal{F}_{20}\cup\\{P_{11}\cup
P_{21},T_{11}\cup T_{21}\\}$
$\displaystyle\cup\left\\{\begin{array}[]{ll}\\{P_{14}\cup P_{24},T_{12}\cup
T_{22}\\}\cup\\{P_{12}\cup P_{23},P_{13}\cup
P_{22},T_{13},T_{14},T_{23},T_{24}\\}&\mbox{if $t=0$};\\\ \\{P_{14}\cup
P_{24},T_{12}\cup T_{22}\\}\cup\\{P_{12},P_{23}\\}\cup\\{P_{13}\cup
P_{22},T_{13},T_{14},T_{23},T_{24}\\}&\mbox{if $t=1$};\\\ \\{P_{14}\cup
P_{24},T_{13}\cup
T_{23}\\}\cup\\{P_{12},P_{22},P_{13},P_{23}\\}\cup\\{T_{12},T_{22},T_{14},T_{24}\\}&\mbox{if
$t=2$};\\\ \\{T_{12}\cup T_{22}\\}\cup\\{P_{12},P_{22},P_{24}\bigtriangleup
B_{s},P_{13},P_{14},P_{23}\\}\cup\\{B_{s},T_{13}\cup P_{24},T_{14}\cup
P_{24}\\}&\mbox{if $t=3$}.\end{array}\right.$

When $|\mathcal{B}_{0}(H_{1})|=0$, let $\mathcal{F}=\mathcal{F}^{*}$. When
$\mathcal{B}_{0}(H_{1})=\\{B_{i}\\}$ for some $i\in[2,h-1]$, let
$\mathcal{C}_{0}=\emptyset$ and

$\mathcal{F}=\left\\{\begin{array}[]{ll}\left(\mathcal{F}^{*}\setminus\\{P_{11}\cup
P_{21},P_{14}\cup P_{24}\\}\right)\cup\\{P_{11}\cup P_{24}\cup
B_{i},P_{14}\cup P_{21}\cup B_{i}\\}&\mbox{if $t\in[0,2]$};\\\
\left(\mathcal{F}^{*}\setminus\\{P_{11}\cup
P_{21},B_{s}\\}\right)\cup\\{B_{s}\cup
T_{1}^{\prime}\\}\cup\\{T_{2}^{\prime}\\}&\mbox{if $t=3$}.\end{array}\right.$

When $|\mathcal{B}_{0}(H_{1})|\geq 2$, let
$\mathcal{F}=\left(\mathcal{F}^{*}\setminus\\{P_{11}\cup
P_{21}\\}\right)\cup\mathcal{C}_{0}\cup\\{T_{1}^{\prime}\cup
T_{2}^{\prime}\\}.$ In each case, one can easily check that $\mathcal{F}$ is a
$\Psi_{xy}(t)$-cover of $H$ by the structure of $H_{2}=D_{1}(x,y)$.

(2) WLOG, assume that $H_{2}=xy$ is positive. Then $h\geq 2$ since $H\not\sim
R_{0}$.

If $\mathcal{B}_{2}(H_{1})=\emptyset$, then $H_{1}=xx_{1}y\cup L_{x_{1}}$ by
Claim 4.1-(6). Thus $H=xx_{1}yx\cup L_{x_{1}}$ is a short barbell by Claim 4.4
and has a $\Psi_{xy}^{*}(2)$-cover.

If $\mathcal{B}_{0}(H_{1})=\emptyset$, then by Claim 4.6-(1), either
$H_{1}\sim D_{2}(x,y)$ in Fig. 2 or $h=2$ and
$\mathcal{B}_{2}(H_{1})=\\{B_{1}\\}$ or $\\{B_{2}\\}$. For the former, $H\sim
R_{3}$ and thus has a $\Psi_{xy}^{*}(2)$-cover. For the latter, by the
symmetry, assume that $\mathcal{B}_{2}(H_{1})=\\{B_{1}\\}$, and then $B_{1}$
has a $\Psi^{*}_{x_{0}x_{1}}(2)$-cover by (1a) and (1b) of Claim 4.6. Since
$H\not\sim R_{i}$ for each $i\in\\{2,4,5\\}$, $H$ has a
$\Psi_{xy}^{*}(2)$-cover by (1c) of Claim 4.6.

Now we assume that $\mathcal{B}_{2}(H_{1})\neq\emptyset$ and
$\mathcal{B}_{0}(H_{1})\neq\emptyset$. Then $h\geq 3$. Let $B_{k}$ (resp.,
$B_{\ell}$) be the part in $\mathcal{B}_{2}(H_{1})\cup\mathcal{B}_{0}(H_{1})$
with minimum (resp., maximum) subscript.

If $V(B_{k})\cap V(B_{\ell})=\emptyset$, then by the choice of
$\mathcal{F}^{*}_{1}$, $V(T_{11})\cap V(T_{13})=\emptyset$. Thus
$T_{11}\cup\\{xy\\}\cup T_{13}$ is a barbell. Therefore, the family

$\mathcal{F}_{10}\cup\mathcal{C}_{0}\cup\\{P_{12}\cup xy,T_{11}\cup xy\cup
T_{13}\\}\cup\\{xy,xy,P_{13},P_{14}\\}\cup\\{T_{1}^{\prime},T_{2}^{\prime},T_{12}\cup
xy,T_{14}\\}$

is a $\Psi_{xy}^{*}(2)$-cover of $H$.

If $V(B_{k})\cap V(B_{\ell})\neq\emptyset$, then either
$\mathcal{B}_{2}(H_{1})=\\{B_{k},B_{k+2}\\}$ and
$\mathcal{B}_{0}(H_{1})=\\{B_{k+1}\\}$, or
$\mathcal{B}_{2}(H_{1})\cup\mathcal{B}_{0}(H_{1})=\\{B_{k},B_{k+1}\\}$. In the
former case, by the proof of Lemma 3.1, there are $4$ negative
$x_{0}x_{h}$-paths
$P_{11}^{\prime},P_{12}^{\prime},P_{13}^{\prime},P_{14}^{\prime}$ in $H_{1}$
such that
$\left(\mathcal{F}_{1}^{*}\setminus\\{P_{11},P_{12},P_{13},P_{14}\\}\right)\cup\\{P_{11}^{\prime},P_{12}^{\prime},P_{13}^{\prime},P_{14}^{\prime}\\}$
is a signed subgraph $6$-cover of $H_{1}$ and hence the family

$\mathcal{F}_{10}\cup\\{P_{11}^{\prime}\cup xy\cup B_{k+1},P_{12}^{\prime}\cup
xy\cup
B_{k+1}\\}\cup\\{xy,xy,P_{13}^{\prime},P_{14}^{\prime}\\}\cup\\{T_{11},xy\cup
T_{13},yx\cup T_{12},T_{14}\\}$

is a $\Psi_{xy}^{*}(2)$-cover of $H$. In the latter case, assume that
$\mathcal{B}_{2}(H_{1})=\\{B_{k}\\}$ and
$\mathcal{B}_{0}(H_{1})=\\{B_{k+1}\\}$ by the symmetry. Then $k=h-2\in[1,2]$
since $H$ has no negative loop at $x_{h}$ and $G$ contains no $2$-vertex. By
Claim 4.6-(2), $B_{k}$ has no $\Psi_{x_{k-1}x_{k}}^{*}(2)$-cover and
$B_{k}\not\sim R_{i}$ for each $i\in\\{2,4,5\\}$. Hence $k=1$; otherwise
$B_{2}\cup B_{1}$ is a piece of $G$ at $\\{x_{2},x_{0}\\}$ and thus, by 1a)
and 1b) of Claim 4.6, $B_{1}$ has a $\Psi_{x_{2}x_{1}}^{*}(2)$-cover, a
contradiction. Since $\mathcal{B}_{0}(H_{1})=\\{B_{2}\\}$ and $H_{2}\cup
H_{3}$ is unbalanced, $G-E(B_{1})$ is coverable. Hence $B_{1}$ is not
coverable. By Claim 4.2, $B_{1}=R_{0}$ and thus $H-L_{x_{1}}\sim R_{2}(y,x)$.
Since $H$ has a unique balanced $3$-circuit $C=x_{0}x_{1}x_{2}x_{0}$, by Claim
4.5, $C\cup H_{3}\sim R_{i}$ for some $i\in\\{2,4,5\\}$. Therefore, one can
easily check that $G=(H-E(C))\cup(C\cup H_{3})$ has a signed circuit
$6$-cover, a contradiction.

(3) Since $xy\notin E(H)$, both $H_{1}$ and $H_{2}$ contain cut vertices by
Claim 4.3, and so $h\geq 2$ and $s-h\geq 2$. If
$|\mathcal{B}_{2}(H_{1})|=|\mathcal{B}_{2}(H_{2})|=0$, then
$H=x_{0}x_{1}x_{2}x_{3}x_{0}\cup\\{L_{x_{1}},L_{x_{3}}\\}$ by Claim 4.1-(6)
and $H-L(H)$ is unbalanced by Claim 4.4. Thus one can easily find a desired
$\Psi_{xy}(2)$-cover, a contradiction. Hence
$|\mathcal{B}_{2}(H_{1})|+|\mathcal{B}_{2}(H_{2})|\geq 1$ and, when
$|\mathcal{B}_{2}(H_{i})|=0$, we may assume that $H_{i}-L(H_{i})$ is positive
(with possible switchings).

By the construction, we can choose $\mathcal{F}_{1}^{*}$ and
$\mathcal{F}_{2}^{*}$ such that $y\notin V(T_{i1})\cap V(T_{i2})$ and $x\notin
V(T_{i3})\cap V(T_{i4})$ for each $i\in[1,2]$; otherwise, if either
$y\in\cup_{i=1}^{2}(V(T_{i1})\cap V(T_{i2}))$ or
$x\in\cup_{i=1}^{2}(V(T_{i3})\cap V(T_{i4}))$, say $y\in V(T_{11})\cap
V(T_{12})$, then
$(\mathcal{B}_{0}(H_{1}),\mathcal{B}_{1}(H_{1}),\mathcal{B}_{2}(H_{1}))=(\emptyset,\\{B_{1}\\},\\{B_{2}\\})$
and for every $\Psi_{x_{1}x_{2}}(2)$-cover of $B_{2}$, both its tadpoles at
$x_{1}$ contain $x_{2}$, contradicting that $B_{2}$ has a
$\Psi_{x_{2}x_{1}}^{*}(2)$-cover by (1a) and (1b) of Claim 4.6. Therefore,
WLOG, assume that $y\notin V(T_{11})\cup V(T_{21})$ and $x\notin V(T_{14})\cup
V(T_{24})$.

If $x\notin V(T_{13})$ or $x\notin V(T_{23})$, say $x\notin V(T_{13})$, since
$|\mathcal{B}_{2}(H_{1})|+|\mathcal{B}_{2}(H_{2})|\geq 1$, the family

$\displaystyle\mathcal{F}$
$\displaystyle=\mathcal{F}_{10}\cup\mathcal{F}_{20}\cup\\{T_{12}\cup
T_{21},T_{13}\cup T_{23}\\}$
$\displaystyle\cup\left\\{\begin{array}[]{ll}\mathcal{C}_{0}\cup\\{T_{11}\cup
T_{22}\\}\cup\mathcal{P}\cup\\{T_{1}^{\prime},T_{2}^{\prime},T_{14},T_{24}\\}&\mbox{if
$\mathcal{B}_{0}(H_{1})\cup\mathcal{B}_{0}(H_{2})\neq\emptyset$};\\\
\\{P_{11}\cup
P_{21}\\}\cup\mathcal{P}\cup\\{T_{11},T_{22},T_{14},T_{24}\\}&\mbox{if
$\mathcal{B}_{0}(H_{1})\cup\mathcal{B}_{0}(H_{2})=\emptyset$}.\end{array}\right.$

is a desired $\Psi_{xy}(2)$-cover, where
$\mathcal{P}=\\{P_{12},P_{13},P_{22},P_{23}\\}\cup\\{P_{14}\cup P_{24}\\}$ if
$|\mathcal{B}_{2}(H_{1})|\geq 1$ and $|\mathcal{B}_{2}(H_{2})|\geq 1$, and
$\mathcal{P}=\\{P_{13},P_{14},P_{23},P_{24}\\}\cup\\{P_{12}\cup P_{22}\\}$
otherwise.

If $x\in V(T_{13})\cap V(T_{23})$, then for each $i\in[1,2]$,
$(\mathcal{B}_{0}(H_{i}),\mathcal{B}_{1}(H_{i}),\mathcal{B}_{2}(H_{i}))=(\emptyset,\\{B_{i+1}\\},\\{B_{3i-2}\\})$,
and both $B_{1}$ and $B_{4}$ have $\Psi_{x_{j-1}x_{j}}^{*}(2)$-covers by (1a)
and (1b) of Claim 4.6. Therefore $H$ has a desired $\Psi_{xy}(2)$-cover by
Claim 3.6-(2). This completes the proof of the claim. ∎

###  4.2 The final step

Since $\overline{G-L(G)}$ is $2$-connected, loopless, $K_{4}$-minor-free, and
of minimum degree at least $3$, it contains a $2$-circuit, denoted by
$C_{1}=x_{0}x_{1}x_{0}$. Let $C_{2}$ be the circuit of $G-L(G)$ corresponding
to $C_{1}$ and let

$B_{1}=C_{2}\cup\\{L_{z}\in L(G):z\in V(C_{2})\setminus\\{x_{0},x_{1}\\}\\}.$

Obviously, $B_{1}$ is a $2$-connected piece of $G$ at $\\{x_{0},x_{1}\\}$. By
Claims 4.4 and 4.1-(6), $C_{2}=B_{1}-L(B_{1})$ is an unbalanced circuit of
length $2$ or $3$ or $4$, denoted by $x_{0}x_{1}x_{0}$ or $x_{0}zx_{1}x_{0}$
or $x_{0}z_{1}x_{1}z_{2}x_{0}$ depending on its length. Hence
$B_{1}=x_{0}x_{1}x_{0}$ or $B_{1}=x_{0}zx_{1}x_{0}\cup L_{z}$ or
$B_{1}=x_{0}z_{1}x_{1}z_{2}x_{0}\cup\\{L_{z_{1}},L_{z_{2}}\\}$. In each case,
$B_{1}$ has a $\Psi_{x_{0}x_{1}}(2)$-cover

$\mathcal{F}_{1}^{*}=\mathcal{F}_{10}\cup\\{P_{11},P_{12},P_{13},P_{14}\\}\cup\\{T_{21},T_{22},T_{23},T_{24}\\},$

where $\mathcal{F}_{10}$ consists of signed circuits, $P_{11}$ and $P_{12}$
(resp., $P_{13}$ and $P_{14}$) are two positive (resp., negative)
$x_{0}x_{1}$-paths, and $T_{11}$ and $T_{12}$ (resp., $T_{13}$ and $T_{14}$)
is two tadpoles at $x_{0}$ (resp., $x_{1}$).

Let $H=H(x_{0},x_{1})$ such that $G=\mathcal{P}(B_{1},H)$. Choose
$B_{i}=B_{i}(x_{i-1},x_{i})$, $i\in[2,s]$, such that

$H(x_{1},x_{0})=\mathcal{S}(B_{2},B_{3},\cdots,B_{s})=B_{2}\cup
B_{3}\cup\cdots\cup B_{s}$

and $s$ is maximum with this property, where $x_{1}\in V(B_{2})$ and
$x_{s}=x_{0}\in V(B_{s})$. Then $|\mathcal{B}_{2}(H)|\geq 1$; otherwise, by
Claim 4.1-(6), $H-L(H)$ is a positive or negative path with length $1$ or $2$,
and thus one can easily find a signed circuit $6$-cover of $G$, a
contradiction. Furthermore, $|\mathcal{B}_{1}(H)|+|\mathcal{B}_{2}(H)|\geq 2$
by Claim 4.3, and every $B_{i}\in\mathcal{B}_{2}(H)$ has a
$\Psi_{x_{i-1}x_{i}}(2)$-cover by Claim 4.7. Applying Lemma 3.1, we pick a
signed subgraph $6$-cover $\mathcal{F}_{2}^{*}$ of $H$ as follows:

$\mathcal{F}_{2}^{*}=\mathcal{F}_{20}\cup
2\mathcal{B}_{0}(H)\cup\\{P_{21},P_{22},P_{23},P_{24}\\}\cup\\{T_{21},T_{22},T_{23},T_{24}\\},$

where $\mathcal{F}_{20}$ is a family of signed circuits, $P_{21}$ and $P_{22}$
(resp., $P_{23}$ and $P_{24}$) are two positive (resp., negative)
$x_{0}x_{1}$-paths, $T_{21}$ and $T_{22}$ (resp., $T_{23}$ and $T_{24}$) are
two tadpoles in $H$ at $x_{0}$ (resp., $x_{1}$) whose unbalanced circuit is in
the part in $\mathcal{B}_{0}(H)\cup\mathcal{B}_{2}(H)$ with maximum (resp.,
minimum) subscript.

Let $U=\bigcap_{B\in\mathcal{B}_{0}(H)\cup\mathcal{B}_{2}(H)}V(B)$. We first
show $U\cap\\{x_{0},x_{1}\\}=\emptyset$. Otherwise $x_{0}\notin V(T_{23})\cap
V(T_{24})$ and $x_{1}\notin V(T_{21})\cap V(T_{22})$. Thus the family

$\displaystyle\mathcal{F}_{10}\cup\mathcal{F}_{20}\cup\\{P_{12}\cup
P_{22},P_{13}\cup P_{23}\\}\cup\\{T_{11}\cup T_{21},T_{12}\cup
T_{22},T_{13}\cup T_{23},T_{14}\cup T_{24}\\}$ $\displaystyle\cup$
$\displaystyle\left\\{\begin{array}[]{ll}\\{P_{14}\cup
P_{24}\\}\cup\mathcal{C}_{0}&\mbox{if }|\mathcal{B}_{0}(H)|\neq 1;\\\
\\{P_{11}\cup P_{24}\cup\mathcal{B}_{0}(H),P_{14}\cup
P_{21}\cup\mathcal{B}_{0}(H)\\}&\mbox{if
$|\mathcal{B}_{0}(H)|=1$}\end{array}\right.$

is a signed circuit $6$-cover of $G$, where $\mathcal{C}_{0}$ is a family of
signed circuits of $P_{11}\cup P_{21}\cup\mathcal{B}_{0}(H)$ which covers
$P_{11}\cup P_{21}$ once and $\mathcal{B}_{0}(H)$ twice, a contradcition.
Hence $U\cap\\{x_{0},x_{1}\\}\neq\emptyset$.

WLOG, assume that $x_{0}\in U$. Then
$\mathcal{B}_{1}(H)=\\{B_{2}\\}=\\{x_{1}x_{2}\\}$,
$\mathcal{B}_{2}(H)=\\{B_{3}\\}$ and
$\mathcal{B}_{0}(H)\in\\{\emptyset,\\{B_{4}\\}\\}$ since
$|\mathcal{B}_{1}(H)|+|\mathcal{B}_{2}(H)|\geq 2$ and
$|\mathcal{B}_{2}(H)|\geq 1$. Hence $G=B_{4}\cup\mathcal{P}(B_{1}\cup
x_{1}x_{2},B_{3})$.

Note that $x_{0}z_{1}x_{1}z_{2}x_{0}\cup\\{L_{z_{1}},L_{z_{2}}\\}$ has a
$\Psi_{x_{0}x_{1}}(2)$-cover in which no tadpole at $x_{1}$ contains $x_{0}$.
Since $B_{1}\cup x_{1}x_{2}$ is a piece of $G$ at $\\{x_{0},x_{2}\\}$, by (1a)
of Claim 4.6, we have either $B_{1}=x_{0}x_{1}x_{0}$ or
$x_{0}zx_{1}x_{0}\cup\\{L_{z}\\}$. Since $B_{3}\cup x_{2}x_{1}$ is a piece of
$G$ at $\\{x_{0},x_{1}\\}$, it follows from Claim 4.7 and (1a) and (1b) of
Claim 4.6 that either $B_{3}$ has a $\Psi_{x_{0}x_{1}}^{*}(2)$-cover or
$B_{3}=B_{3}(x_{0},x_{2})\sim R_{i}(x,y)$ for some $i\in\\{0,2,4,5\\}$.
Therefore, by Lemma 3.3-(2), $G$ has a signed circuit $6$-cover, a
contradiction. This completes the proof of Theorem 1.2.

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