Removable Cycles Avoiding Two Connected Subgraphs

Abstract

We provide a sufficient condition for the existence of a cycle 𝐶 in a connected graph 𝐺 which is edge-disjoint from two connected subgraphs 𝐻₁ and 𝐻₂ of 𝐺 such that 𝐺−𝐸(𝐶) is connected.

1. Introduction

Prompted by Hobbs' conjecture, Jackson [1] proved that if 𝐺 is a 2-connected simple graph of minimum degree at least 4 and 𝑒∈𝐸(𝐺), then 𝐺−𝑒 contains a cycle 𝐶 such that 𝐺−𝐸(𝐶) is 2-connected. Lemos and Oxley [2] generalized this result as follows.

Theorem 1. Let 𝐺 be a 2-connected simple graph and let 𝐻 be a subgraph of 𝐺 such that 𝐻 is either 2-connected or 𝐾₂. Suppose that 𝑑_𝐺(𝑣)≥4 for all 𝑣 in 𝑉(𝐺)−𝑉(𝐻) and 𝐺 has a cycle that is edge-disjoint from 𝐻. Then there exists a cycle 𝐶 in 𝐺 that is edge-disjoint from 𝐻 such that 𝐺−𝐸(𝐶) is 2-connected.

Borse and Waphare [3] obtained the following result for the class of connected graphs which is analogous to the above theorem and is an improvement of a result due to Sinclair [4].

Theorem 2. Let 𝐺 be a connected simple graph and let 𝐻 be a connected subgraph of 𝐺 such that there is a cycle in 𝐺 that is edge-disjoint from 𝐻. Suppose that 𝑑_𝐺(𝑣)≥3 for all 𝑣∈𝑉(𝐺)−𝑉(𝐻). Then there exists a cycle 𝐶 in 𝐺 that is edge-disjoint from 𝐻 such that 𝐺−𝐸(𝐶) is connected.

The problem of the existence of cycles in graphs deletion of whose edges preserve connectedness is well studied in the literature (see [1–10]). In this paper, we improve Theorem 2 as follows.

Theorem 3. Let 𝐺 be a connected simple graph and let 𝐻₁ and 𝐻₂ be connected subgraphs of 𝐺. Suppose that there are at least two cycles in 𝐺−𝐸(𝐻₁∪𝐻₂) and further, no two edges of 𝐺−𝐸(𝐻₁∪𝐻₂) form an edge-cut of 𝐺. Then there exists a cycle 𝐶 in 𝐺 that is edge-disjoint from both 𝐻₁ and 𝐻₂ such that 𝐺−𝐸(𝐶) is connected.

Let 𝐺 be a connected graph of minimum degree at least three and let 𝐻₁ and 𝐻₂ be connected subgraphs of 𝐺. If 𝐺−𝐸(𝐻₁∪𝐻₂) has only one cycle, say 𝐶, then 𝐺−𝐸(𝐶) is not connected if a subset of 𝐸(𝐶) forms an edge-cut of 𝐺. Therefore, in Theorem 3, we must assume that 𝐺−𝐸(𝐻₁∪𝐻₂) contains at least two cycles. The following example shows that the condition in Theorem 3 regarding an edge-cut of 𝐺 is necessary. Let 𝐺₁ and 𝐺₂ be two disjoint copies of 𝐾_𝑛 with 𝑛≥4. Suppose that 𝑒_𝑖=𝑢_𝑖𝑣_𝑖 is an edge of 𝐺_𝑖 for 𝑖=1,2. Let 𝐻_𝑖=𝐺_𝑖−𝑒_𝑖 for 𝑖=1,2. Let 𝐺 be the graph obtained from 𝐻₁ and 𝐻₂ by adding two new vertices 𝑢, 𝑣 and seven new edges 𝑢₁𝑢, 𝑢𝑢₂, 𝑢₂𝑣₂, 𝑣₂𝑣, 𝑣𝑣₁, 𝑣₁𝑢₁, and 𝑢𝑣; see Figure 1. Then 𝐺 is simple, connected and the minimum degree of 𝐺 is 3. Let 𝐶₁=𝑢₁𝑢+𝑢𝑣+𝑣𝑣₁+𝑣₁𝑢₁,  𝐶₂=𝑢𝑢₂+𝑢₂𝑣₂+𝑣₂𝑣+𝑣𝑢, and C₃=𝑢₁𝑢+𝑢𝑢₂+𝑢₂𝑣₂+𝑣₂𝑣+𝑣𝑣₁+𝑣₁𝑢₁. Then 𝐶₁,𝐶₂, and 𝐶₃ are the only cycles of 𝐺−𝐸(𝐻₁∪𝐻₂). For 𝐶∈{𝐶₁,𝐶₂,𝐶₃}, it is easy to see that 𝐺−𝐸(𝐶) is not connected as the set {𝑢𝑢_𝑖,𝑣𝑣_𝑖} is a 2-edge-cut of 𝐺 for each 𝑖∈{1,2}.

Figure 1 

As a consequence of Theorem 3, we get the following result.

Corollary 4. Let 𝐺 be an Eulerian 3-edge-connected simple graph containing at least four mutually edge-disjoint cycles. Then there exist three mutually edge-disjoint cycles 𝐶₁,𝐶₂, and 𝐶₃ such that 𝐺−𝐸(𝐶_𝑖) is connected for 𝑖=1,2,3.

We refer to [11] for the terms that are not defined here.

2. Proof

In this section, we prove Theorem 3 and its corollary. First we prove the following lemma.

Lemma 5. Let 𝐺,𝐻₁, and 𝐻₂ be as in the hypothesis of Theorem 3. If there is an edge of 𝐺 joining 𝐻₁ and 𝐻₂, then there exists a cycle 𝐶 in 𝐺 that is edge-disjoint from both 𝐻₁ and 𝐻₂ such that 𝐺−𝐸(𝐶) is connected.

Proof. Let 𝑒 be an edge of 𝐺, having one end vertex in 𝐻₁ and the other end vertex in 𝐻₂. Let 𝐻^′=𝐻₁∪𝐻₂∪𝑒. Then 𝐻^′ is a connected subgraph of 𝐺. Let 𝐶₁ and 𝐶₂ be cycles in 𝐺−𝐸(𝐻₁∪𝐻₂). If 𝐶₁ or 𝐶₂ does not contain 𝑒, then it is a cycle in 𝐺−𝐸(𝐻^′). If 𝑒 belongs to both 𝐶₁ and 𝐶₂, then (𝐶₁∪𝐶₂)−𝑒 contains a cycle. Thus, in any case, there is a cycle in 𝐺−𝐸(𝐻^′). Since 𝐺−𝑋 is connected for every subset 𝑋 of edges of the graph 𝐺−𝐸(𝐻^′) with |𝑋|≤2,  𝑑_𝐺(𝑣)≥3 for all 𝑣∈𝑉(𝐺)−𝑉(𝐻^′). Hence, by Theorem 2, there exists a cycle 𝐶 in 𝐺−𝐸(𝐻^′) such that 𝐺−𝐸(𝐶) is connected. Obviously, 𝐶 is edge-disjoint from both 𝐻₁ and 𝐻₂.

The number of edges of a cycle or a path 𝑃 is denoted by |𝑃|. If 𝑥 and 𝑦 are distinct vertices of a cycle 𝐶 of a graph 𝐺, then 𝐶[𝑥,𝑦] and 𝐶[𝑦,𝑥] denote the edge-disjoint paths along 𝐶 which join 𝑥 to 𝑦.

Proof of Theorem 3. Suppose that the result is false and let 𝐺 be a counterexample. Then there exist connected subgraphs 𝐻₁ and 𝐻₂ of 𝐺 satisfying the hypothesis of the theorem such that for every cycle 𝐶 in 𝐺 that is edge-disjoint from 𝐻₁∪𝐻₂, the graph 𝐺−𝐸(𝐶) is not connected. Choose 𝐻₁ and 𝐻₂ such that |𝐸(𝐻₁∪𝐻₂)| is maximum. By a hypothesis, 𝑑_𝐺(𝑣)≥3 for all 𝑣∈𝑉(𝐺)−𝑉(𝐻₁∪𝐻₂). Therefore, by Theorem 2, 𝐻₁∪𝐻₂ is not connected. Thus 𝐻₁ is vertex disjoint from 𝐻₂.
Claim 1. If 𝐶 is a cycle in 𝐺−𝐸(𝐻₁∪𝐻₂), then every component of 𝐺−𝐸(𝐶) contains 𝐻₁ or 𝐻₂.
Let 𝐶 be a cycle in 𝐺−𝐸(𝐻₁∪𝐻₂). Since 𝐻_𝑖 is connected, it is contained in a component 𝐷_𝑖 of 𝐺−𝐸(𝐶) for 𝑖=1,2. If 𝐷₁=𝐷₂; then, by Theorem 2, there exists a cycle 𝑍 in 𝐺−𝐸(𝐷₁) such that 𝐺−𝐸(𝑍) is connected, which is a contradiction. Therefore 𝐷₁≠𝐷₂. We prove that 𝐷₁ and 𝐷₂ are the only components of 𝐺−𝐸(𝐶). Let 𝐴=𝑉(𝐶)−𝑉(𝐷₁∪𝐷₂). It is sufficient to prove that 𝐴 is an empty set. Assume that 𝐴 is nonempty. Since 𝐺 is connected, each component of 𝐺−𝐸(𝐶) contains a vertex of 𝐶. If a component of 𝐺−𝐸(𝐶) contains exactly one vertex of 𝐶, then the edges incident with that vertex form a 2-edge-cut of 𝐺, which is a contradiction. Hence each component of 𝐺−𝐸(𝐶) contains at least two vertices of 𝐶. Hence |𝐴|≥2. Therefore there is a component 𝐷 of 𝐺−𝐸(𝐶) different from 𝐷₁ and 𝐷₂ such that |𝑉(𝐷)∩𝐴|≥2. Let 𝑥₁,𝑥₂∈𝑉(𝐷)∩𝐴. Let 𝑃 be a path in 𝐷 between two vertices of 𝐶. Then 𝑃∪𝐶 contains at least two cycles which are edge-disjoint with 𝐷₁∪𝐷₂. By the maximality of |𝐸(𝐻₁∪𝐻₂)|, we have 𝐷₁=𝐻₁ and 𝐷₂=𝐻₂. By Lemma 5, there is no edge between 𝐷₁ and 𝐷₂. Suppose that |𝐴|=2. Then 𝐴={𝑥₁,𝑥₂}. It follows that 𝐺−𝐸(𝐶) has exactly three components and hence we get two edges 𝑒₁,𝑒₂ of 𝐶, with each having one end in 𝐷₁ and the other in 𝐷 such that {𝑒₁,𝑒₂} is an edge-cut of 𝐺, a contradiction. Thus |𝐴|≥3.
Let 𝑤∈𝑉(𝐶)∩𝑉(𝐷₁). Then 𝑤∉𝐴. By traversing from 𝑤 along 𝐶 on both sides, let 𝑣₁∉𝑉(𝐷₁) be the first vertex of 𝐶 on one side and let 𝑣₂∉𝑉(𝐷₁) be the first vertex of 𝐶 on the other side. By Lemma 5, 𝑣₁,𝑣₂∉𝑉(𝐷₂). We may assume that 𝑤∈𝑉(𝐶[𝑣₁,𝑣₂]). Then 𝑉(𝐶[𝑣₂,𝑣₁])∩𝐴=𝐴. Let 𝐷^′₁=𝐷₁∪𝐶[𝑣₁,𝑣₂]. Then 𝐷^′₁ is connected and |𝐸(𝐷^′₁∪𝐷₂)|>|𝐸(𝐻₁∪𝐻₂)|. By the maximality of |𝐸(𝐻₁∪𝐻₂)|,  𝐺−𝐸(𝐷^′₁∪𝐷₂) contains at most one cycle. Let 𝐷₃ be the component of 𝐺−𝐸(𝐶) containing 𝑣₂. Then 𝐷₃∉{𝐷₁,𝐷₂} and |𝑉(𝐷₃)∩𝐴|≥2. If |𝑉(𝐷₃)∩𝐴|≥3, then 𝐷₃∪𝐶[𝑣₂,𝑣₁] contains two cycles, which is a contradiction. Hence |𝑉(𝐷₃)∩𝐴|=2. This implies that there is a component 𝐷₄ of 𝐺−𝐸(𝐶) such that 𝐷₄∉{𝐷₁,𝐷₂,𝐷₃}. Traverse the path 𝐶[𝑣₂,𝑣₁] beginning at 𝑣₂, let 𝑣^′₂≠𝑣₂ be the second vertex of 𝐷₃ on this path. Let 𝑃_{𝑣2𝑣′2} be a 𝑣₂,𝑣^′₂ path in 𝐷₃. Let 𝐶₁=𝑃_{𝑣2𝑣′2}∪𝐶[𝑣₂,𝑣^′₂]. Then 𝐶₁ is a cycle. If 𝐶[𝑣₂,𝑣^′₂] does not meet 𝐷₄, then 𝐶[𝑣^′₂,𝑣₁] intersects 𝐷₄ in at least two vertices giving a cycle which is disjoint from 𝐶₁, a contradiction. Hence 𝐶[𝑣₂,𝑣^′₂] contains a vertex 𝑥 of 𝐷₄. If this path contains another vertex 𝑦 of 𝐷₄, then 𝐶₁∪𝑃_𝑥𝑦 contains two cycles, where 𝑃_𝑥𝑦 is an 𝑥,𝑦 path in 𝐷₄. This is a contradiction. Hence 𝑉(𝐶[𝑣₂,𝑣^′₂])∩𝑉(𝐷₄)={𝑥}. This implies that the path 𝐶[𝑣^′₂,𝑣₁] meets 𝐷₄ in a vertex 𝑧≠𝑥. Let 𝑃_𝑥𝑧 be an 𝑥,𝑧 path in 𝐷₄. Then 𝐶₁∪𝐶[𝑣^′₂,𝑧]∪𝑃_𝑥𝑧 contains two cycles, which is a contradiction. This proves the claim.
Claim 2. If 𝐶₁ and 𝐶₂ are cycles in 𝐺−𝐸(𝐻₁∪𝐻₂), then |𝑉(𝐶₁)∩𝑉(𝐶₂)|≥2.
Assume that the claim is false. Let 𝐶₁ and 𝐶₂ be cycles in 𝐺−𝐸(𝐻₁∪𝐻₂) such that |𝑉(𝐶₁)∩𝑉(𝐶₂)|<2. Then 𝐶₁ is edge-disjoint from 𝐶₂. By Claim 1, each component of 𝐺−𝐸(𝐶₁) contains 𝐻₁ or 𝐻₂. Thus 𝐺−𝐸(𝐶₁) has exactly two components. We may assume that 𝐻_𝑖 is contained in a component 𝐷_𝑖 of 𝐺−𝐸(𝐶₁) for 𝑖=1,2. Obviously, 𝐶₂ is contained in 𝐷₁ or 𝐷₂. We may assume that 𝐶₂ is contained in 𝐷₁. By the maximality of |𝐸(𝐻₁∪𝐻₂)|, it follows that 𝐻₂=𝐷₂. By Claim 1, 𝐺−𝐸(𝐶₂) has only two components 𝐷^′₁ and 𝐷^′₂, with each containing 𝐻₁ or 𝐻₂. We may assume that 𝐻_𝑖 is contained in 𝐷^′_𝑖 for 𝑖=1,2. Since 𝐻₂∪𝐶₁ is connected, it is contained in 𝐷^′₂. Let 𝑄 be the set of all edges in 𝐺, having one end vertex in 𝐷₁ and the other end vertex in 𝐷₂. Then 𝑄⊆𝐸(𝐶₁). By a hypothesis, |𝑄|≥4. Let 𝑄^′ be the set of all edges in 𝐺, having one end vertex in 𝐷^′₁ and the other end vertex in 𝐷^′₂. Then 𝑄^′⊆𝐸(𝐶₂) and |𝑄^′|≥4.
If 𝐷₁−𝐸(𝐶₂) is connected, then 𝐺−𝐸(𝐶₂) is connected, a contradiction. Therefore we may assume that 𝐷₁−𝐸(𝐶₂) is disconnected. Obviously, 𝐷^′₁ is a component of 𝐷₁−𝐸(𝐶₂). Let 𝑆=𝑉(𝐺)−(𝑉(𝐷₂)∪𝑉(𝐷^′₁)). Note that |𝑉(𝐶₁)∩𝑉(𝐷₁)|≥2,  |𝑉(𝐶₁)∩𝑉(𝐷^′₁)|=0,  |𝑉(𝐶₂)∩𝑉(𝐷^′₂)|≥2, and |𝑉(𝐶₁)∩𝑉(𝐶₂)|≤1. Hence |𝑆|≥3. Let 𝐻=𝐺[𝑆]. Then 𝐻 is vertex disjoint from both 𝐻₁ and 𝐻₂. If 𝐻 contains a cycle 𝐶^′ then 𝐶₁∪𝐶^′ contains two cycles and hence we get two cycles in 𝐺−𝐸(𝐷₂∪(𝐷^′₁∪𝑄^′)). By the maximality of |𝐸(𝐻₁∪𝐻₂)|, we get a cycle 𝐶 that is edge-disjoint from both 𝐻₁ and 𝐻₂ such that 𝐺−𝐸(𝐶) is connected, which is a contradiction. Thus 𝐻 is a forest. Let 𝐻^′=𝐻−𝐸(𝐶₁∪𝐶₂). Then 𝐻^′ is a forest. Every edge 𝑒 of 𝐺 with one end vertex in 𝑆 and the other end vertex in 𝑉(𝐺)−𝑆 is either in 𝑄 or in 𝑄^′ and hence 𝑒∈𝐸(𝐶₁) or 𝑒∈𝐸(𝐶₂). As 𝑑_𝐺(𝑣)≥3 for all 𝑣∈𝑉(𝐻^′) and |𝑉(𝐶₁)∩𝑉(𝐶₂)|≤1, at most one vertex of 𝐻^′ is isolated. Hence 𝐻^′ contains an edge. Let 𝑇 be a nontrivial tree in 𝐻^′. Let 𝑣₁ and 𝑣₂ be two pendant vertices of 𝑇. Since 𝑑_𝐺(𝑣_𝑖)≥3, 𝑣_𝑖∈𝑉(𝐶₁)∪𝑉(𝐶₂) for 𝑖=1,2. We may assume that 𝑣₁∈𝑉(𝐶₁). Suppose that |𝑉(𝐶₁)∩𝑉(𝑇)|≥2. Let {𝑥,𝑦}⊆𝑉(𝑇)∩𝑉(𝐶₁) and let 𝑃 be the 𝑥,𝑦 path in 𝑇. Then 𝐶₁∪𝑃 contains at least two cycles. These cycles are edge-disjoint from 𝐷^′₁∪𝑄^′ and 𝐷₂. Now, a contradiction follows from the maximality of |𝐸(𝐻₁∪𝐻₂)|. Hence |𝑉(𝑇)∩𝑉(𝐶₁)|≤1. Similarly, |𝑉(𝑇)∩𝑉(𝐶₂)|≤1. This implies that 𝑉(𝐶_𝑖)∩𝑉(𝑇)={𝑣_𝑖} for 𝑖=1,2. If 𝑇 has a third pendant vertex 𝑣₃, then 𝑣₃ is in 𝑉(𝐶₁) or 𝑉(𝐶₂), which is a contradiction. Hence 𝑇 has exactly two pendant vertices and therefore it is a path. As 𝑑_𝐺(𝑣)≥3 for all 𝑣∈𝑉(𝑇),  𝑉(𝑇)⊆𝑉(𝐶₁)∪𝑉(𝐶₂). This implies that 𝑇 has no internal vertex. Hence 𝑇 is isomorphic to 𝐾₂. Thus every component of 𝐻^′ is isomorphic to either 𝐾₁ or 𝐾₂. Let 𝑒₁=𝑢₁𝑢₂ be an edge of 𝐻^′ with 𝑢₁∈𝑉(𝐶₁) and 𝑢₂∈𝑉(𝐶₂). Suppose that 𝐻^′ has an another edge 𝑒₂=𝑢₃𝑢₄. Then 𝑑_𝐻′(𝑢_𝑖)=1 for each 𝑖∈{1,2,3,4}. We may assume that 𝑢₃∈𝑉(𝐶₁) and 𝑢₄∈𝑉(𝐶₂). Since 𝑄^′⊂𝐸(𝐶₂), we may assume by suitable labeling that at least one edge of 𝑄^′ is not in 𝐶₂[𝑢₂,𝑢₄]. It follows that the subgraph 𝐶₁∪𝐶₂[𝑢₂,𝑢₄]∪𝑒₁∪𝑒₂ is edge-disjoint from the subgraph 𝐷₂∪𝐷^′₁∪𝐺[𝑄^′−𝐸(𝐶₂[𝑢₂,𝑢₄])] and further, it contains at least two cycles. Obviously, 𝐷^′₁∪𝐺[𝑄^′−𝐸(𝐶₂[𝑢₂,𝑢₄])] is connected and contains more edges than 𝐻₁. A contradiction follows from the maximality of |𝐸(𝐻₁∪𝐻₂)|. Hence 𝐻^′ contains exactly one edge. Since 𝑉(𝐻)=𝑉(𝐻^′),  |𝑉(𝐻^′)|≥3. As at most one vertex of 𝐻^′ is isolated, |𝑉(𝐻)|=3. Let 𝑉(𝐻)={𝑢₁,𝑢₂,𝑢₃}. We may assume that 𝑢₁𝑢₂∈𝐸(𝐺),  𝑢₁∈𝑉(𝐶₁),  𝑢₂∈𝑉(𝐶₂), and 𝑢₃∈𝑉(𝐶₁)∩𝑉(𝐶₂). Obviously, 𝑄^′−𝐸(𝐶₂[𝑢₂,𝑢₃])≠𝜙 or 𝑄^′−𝐸(𝐶₂[𝑢₃,𝑢₂])≠𝜙. We may assume that 𝑄^′−𝐸(𝐶₂[𝑢₂,𝑢₃])≠𝜙. The subgraph 𝐶₂[𝑢₂,𝑢₃]∪𝑢₁𝑢₂∪𝐶₁ contains two cycles and further, it is edge-disjoint from the subgraphs 𝐷₂ and 𝐷^′₁∪𝐺[𝑄^′−𝐸(𝐶₂[𝑢₂,𝑢₃])]. A contradiction follows from the maximality of |𝐸(𝐻₁∪𝐻₂)|. The proof of Claim 2 is completed.
By Claim 2, any two cycles in 𝐺−𝐸(𝐻₁∪𝐻₂) has at least two vertices in common. This implies that there exist two vertices 𝑣₁ and 𝑣₂ and three mutually internally disjoint 𝑣₁, 𝑣₂ paths 𝑃₁, 𝑃₂, and 𝑃₃ in 𝐺−𝐸(𝐻₁∪𝐻₂). We may assume that |𝑃₁|≤|𝑃₂|≤|𝑃₃|. Let 𝐶₁=𝑃₁+𝑃₂. Then 𝐶₁ is a cycle in 𝐺−𝐸(𝐻₁∪𝐻₂). By Claim 1, 𝐺−𝐸(𝐶₁) has exactly two components each containing 𝐻₁ or 𝐻₂. Let 𝐷_𝑖 be the component of 𝐺−𝐸(𝐶₁) containing 𝐻_𝑖 for 𝑖=1,2. We may assume that the path 𝑃₃ is contained in the component 𝐷₁. Hence the vertices 𝑣₁ and 𝑣₂ are in 𝐷₁. It follows from the maximality of |𝐸(𝐻₁∪𝐻₂)| that 𝐷₂=𝐻₂. Let 𝑄 be the set of all edges in 𝐺 that have one end vertex in 𝐷₁ and the other end vertex in 𝐷₂. Then 𝑄⊆𝐸(𝐶₁). By a hypothesis, |𝑄|≥3. In fact, |𝑄|≥4. By Lemma 5, no edge of 𝑄 has an end vertex in 𝐻₁ as 𝐻₂=𝐷₂. Obviously, 𝐸(𝑃₁)∩𝑄≠𝜙 or 𝐸(𝑃₂)∩𝑄≠𝜙. Let 𝐶^′=𝑃₂+𝑃₃ if 𝐸(𝑃₁)∩𝑄≠𝜙; otherwise 𝐶^′=𝑃₁+𝑃₃. Thus 𝐶^′ is a cycle in 𝐺 that is edge-disjoint from both 𝐻₁ and 𝐻₂ and further, 𝐸(𝑃₃)⊆𝐸(𝐶^′) and 𝐸(𝑃^′)∩𝐸(𝐶^′)=𝜙 for some 𝑃^′∈{𝑃₁,𝑃₂}. By Claim 1, 𝐺−𝐸(𝐶^′) has exactly two components 𝐷^′₁ and 𝐷^′₂. We may assume that 𝐻₁ is contained in 𝐷^′₁ and 𝐷₂ is contained in 𝐷^′₂. As 𝐸(𝑃^′)∩𝑄≠𝜙,  𝐷₂∪𝑃^′ is connected and hence 𝐷^′₂ contains 𝑃^′. So the vertices 𝑣₁ and 𝑣₂ do not belong to 𝐷^′₁. Thus 𝐷₁−𝐸(𝑃₃) has at least two components, one of which is 𝐷^′₁.
By a hypothesis, |𝑉(𝐷^′₁)∩𝑉(𝐶^′)|≥2. Further, every component of 𝐷₁−𝐸(𝑃₃) contains at least one vertex of 𝑃₃ and hence contains at least one vertex of 𝐶^′. Suppose that there is a component 𝐹 of 𝐷₁−𝐸(𝑃₃) containing two vertices 𝑥₁ and 𝑥₂ of 𝐶^′ with 𝐹≠𝐷^′₁. Let 𝑃_𝑥1𝑥2 be an 𝑥₁,𝑥₂ path in 𝐹. Obviously, 𝑃_𝑥1𝑥2∪𝐶^′ contains at least two cycles. Thus there are at least two cycles in 𝐺−𝐸(𝐷^′₁∪(𝑃^′∪𝐷₂)). Further, 𝐸(𝐻₁∪𝐻₂)⊂𝐸(𝐷^′₁∪(𝑃^′∪𝐷₂)). Since 𝐷^′₁ and 𝐷₂∪𝑃^′ are connected and are vertex disjoint subgraphs, by the maximality of |𝐸(𝐻₁∪𝐻₂)|, there exists a cycle 𝐶 in 𝐺 that is edge-disjoint from 𝐷^′₁∪(𝑃^′∪𝐷₂) and hence edge-disjoint from 𝐻₁∪𝐻₂ such that 𝐺−𝐸(𝐶) is connected, which is a contradiction. Thus 𝐶^′ has exactly one vertex in common with every component of 𝐷₁−𝐸(𝑃₃) that is different from 𝐷^′₁. Since 𝑣₁ and 𝑣₂ are common vertices of 𝐷₁−𝐸(𝑃₃) and 𝐶^′, there are two components 𝐹₁ and 𝐹₂ of 𝐷₁−𝐸(𝑃₃), both different from 𝐷^′₁ such that 𝑣₁∈𝑉(𝐹₁) and 𝑣₂∈𝑉(𝐹₂).
Suppose that 𝐹₁, 𝐹₂, and 𝐷^′₁ are the only components of 𝐷₁−𝐸(𝑃₃). Let 𝑃^′′∈{𝑃₁,𝑃₂}−{𝑃^′}. Then 𝐶^′=𝑃₃+𝑃^′′. Suppose that 𝑃^′′ does not meet 𝐷^′₁. Then, as 𝑃₃ meets each of 𝐹₁ and 𝐹₂ only once, the corresponding two edges of 𝑃₃ form a 2-edge-cut of 𝐺, which is a contradiction. Therefore 𝑃^′′ meets 𝐷^′₁. Since 𝑃^′′ is a 𝑣₁,𝑣₂ path and |𝑉(𝐶^′)∩𝑉(𝐹_𝑖)|=1,𝑉(𝑃^′′)∩𝑉(𝐹_𝑖)={𝑣_𝑖} for 𝑖=1,2. If 𝑃^′′ meets 𝐷₂, then 𝐷^′₁∪𝐷₂∪𝐹₁∪𝐹₂∪𝑃^′′=𝐺−𝐸(𝑃₃+𝑃^′) is connected, which is a contradiction as 𝑃₃∪𝑃^′ is a cycle in 𝐺−𝐸(𝐻₁∪𝐻₂). Thus 𝑃^′′ does not meet 𝐷₂. Hence 𝑄∩𝑃^′′=𝜙. This shows that 𝑄⊆𝐸(𝑃^′). Traverse the path 𝑃^′ beginning at 𝑣₁ and suppose that 𝑃^′ meets 𝐹₂ at the first time in the vertex 𝑥. Let 𝑦 be the last vertex of 𝐹₁ on the 𝑣₁,𝑥 subpath 𝑃_𝑣1𝑥 of 𝑃^′. Then the subpath 𝑃_𝑥𝑦 of 𝑃^′ contains at most two edges of 𝑄. Extend the path 𝑃_𝑥𝑦 to a 𝑣₁,𝑣₂ path 𝑃_𝑣1𝑣2 by adding a 𝑦,𝑣₁ path in 𝐹₁ and an 𝑥,𝑣₂ path in 𝐹₂. Thus 𝑄−𝐸(𝑃_𝑣1𝑣2)≠𝜙. As 𝐶^′∪𝑃_𝑣1𝑣2 contains at least two cycles, there are at least two cycles that are edge-disjoint from the connected subgraphs 𝐷^′₁ and 𝐷₂∪𝐺[𝑄−𝐸(𝑃_𝑣1𝑣2)] of 𝐺. Since 𝐻₁ is contained in 𝐷^′₁,  𝐻₂ is contained in 𝐷₂, and 𝐷₂∪𝐺[𝑄−𝐸(𝑃_𝑣1𝑣2)] is connected, by the maximality of |𝐸(𝐻₁∪𝐻₂)|, there exists a cycle 𝐶 that is edge-disjoint from 𝐷^′₁∪𝐷₂∪𝐺[𝑄−𝐸(𝑃_𝑣1𝑣2)] such that 𝐺−𝐸(𝐶) is connected, which is a contradiction. Thus 𝐷₁−𝐸(𝑃₃) has at least four components.
Since 𝐷^′₁ and 𝐷^′₂ are the only components of 𝐺−𝐸(𝐶^′), the path 𝑃^′ meets every component of 𝐷₁−𝐸(𝑃₃) except 𝐷^′₁. Also, |𝑄∩𝐸(𝑃^′)|≥2. It follows that we can choose two components 𝐹^′₁ and 𝐹^′₂ of 𝐷₁−𝐸(𝑃₃) and two vertices 𝑥∈𝑉(𝐹^′₁)∩𝑉(𝑃^′),  𝑦∈𝑉(𝐹^′₂)∩𝑉(𝑃^′) such that for the 𝑥,𝑦 subpath 𝑃_𝑥𝑦 of 𝑃^′ we have (𝐸(𝑃^′)∩𝑄)−𝐸(𝑃_𝑥𝑦)≠𝜙. Let 𝑥^′∈𝑉(𝐶^′)∩𝑉(𝐹^′₁),𝑦^′∈𝑉(𝐶^′)∩𝑉(𝐹^′₂) and let 𝑃_𝑥𝑥′ and 𝑃_𝑦𝑦′ be paths in 𝐹^′₁ and 𝐹^′₂, respectively, from 𝑥 to 𝑥^′ and 𝑦 to 𝑦^′. Then 𝐶^′∪(𝑃_𝑥′𝑥+𝑃_𝑥𝑦+𝑃_𝑦𝑦′) contains at least two cycles and further, it is edge-disjoint from 𝐷^′₁,𝐷₂, and (𝐸(𝑃^′)∩𝑄)−𝐸(𝑃_𝑥𝑦). Due to the maximality of |𝐸(𝐻₁∪𝐻₂)|, there exists a cycle 𝐶 that is edge-disjoint from both 𝐷₁ and 𝐷₂∪𝐺[(𝐸(𝑃^′)∩𝑄)−𝐸(𝑃_𝑥𝑦)] such that 𝐺−𝐸(𝐶) is connected, which is a contradiction. This completes the proof of the theorem.

Proof of Corollary 4. Let 𝑍₁, 𝑍₂, 𝑍₃ and 𝑍₄ be mutually edge-disjoint cycles in 𝐺. By Theorem 3, there exists a cycle 𝐶₁ in 𝐺−𝐸(𝑍₁∪𝑍₂) such that 𝐺−𝐸(𝐶₁) is connected. Since 𝐺 is eulerian, every component of 𝐺−𝐸(𝑍₁∪𝑍₂) is eulerian. Hence 𝐺−𝐸(𝑍₁∪𝑍₂) has a cycle decomposition ℒ₁ containing the cycle 𝐶₁. Since 𝑍₃ and 𝑍₄ are edge-disjoint from 𝑍₁ and 𝑍₂,ℒ₁ contains at least one cycle 𝐶^′₁ other than 𝐶₁. As 𝐶₁∪𝐶^′₁ is edge-disjoint from 𝑍₁∪𝑍₂, by Theorem 3, there exists a cycle 𝐶₂ in 𝐺−𝐸(𝐶₁∪𝐶^′₁) such that 𝐺−𝐸(𝐶₂) is connected. Obviously, each component of 𝐺−𝐸(𝐶₁∪𝐶₂) is eulerian. Further, 𝐶^′₁ is a cycle in 𝐺−𝐸(𝐶₁∪𝐶₂). There exists a cycle decomposition ℒ₂ of 𝐺−𝐸(𝐶₁∪𝐶₂) containing 𝐶^′₁. It is easy to see that 𝐺−𝐸(𝐶₁∪𝐶^′₁∪𝐶₂) contains at least one edge. Hence 𝐺−𝐸(𝐶₁∪𝐶₂) contains at least one edge which is not in 𝐶^′₁. This implies that ℒ₂ contains a cycle 𝐶^′₂ with 𝐶^′₂≠𝐶^′₁. Hence, by Theorem 3, there exists a cycle 𝐶₃ in 𝐺−𝐸(𝐶₁∪𝐶₂) such that 𝐺−𝐸(𝐶₃) is connected. Obviously, 𝐶₁,  𝐶₂, and 𝐶₃ are mutually edge-disjoint cycles.