UMBC CMSC203, Discrete Structures, Fall 2010

Equivalence Proofs

G has an Euler circuit ⇐⇒ every vertex in G has even degree.

G has an Euler circuit ⇒ every vertex in G has even degree.

Every vertex in G has even degree ⇒ G has an Euler circuit.

Another common pitfall in equivalence proofs is to prove the same direction twice. Since we can use a direct or an indirect proof for either the "if" or the "only if" direction, there are 4 possible ways we can write an equivalence proof. To prove that p ⇐⇒ q, we can use any of the following 4 methods:

1. Show p ⇒ q and q ⇒ p .
2. Show p ⇒ q and ¬p ⇒ ¬q.
3. Show ¬q ⇒ ¬p and q ⇒ p .
4. Show ¬q ⇒ ¬p and ¬p ⇒ ¬q .

p ⇒ q   and   ¬q ⇒ ¬p

Sometimes a theorem establishes the equivalence of several propositions p₁, p₂, p₃, ..., p_n. In this case, we do not need to prove the equivalence between every pair of statements. All we need to show is:

p₁ ⇒ p₂
p₂ ⇒ p₃
p₃ ⇒ p₄
...
p_n-1 ⇒ p_n
p_n ⇒ p₁

Claim: Let G be a connected undirected graph with n vertices and e edges. Then, the following statements are equivalent:

1. G does not contain any cycles.
2. e = n − 1.
3. there is a unique path between every pair of vertices in G.

Proof:
(I ⇒ II)   Suppose that G does not contain any cycles. We first argue that G must have a vertex with degree 1 as long as G has at least 2 vertices. (We do this because when we remove a degree 1 vertex, we do not disconnect the graph and we reduce the number of vertices and edges by 1.) To see this, we explore the graph along a path. We start at any vertex and move along an edge to another vertex. We keep moving along edges but we do not reuse the edge that we used to arrive at the current vertex. If we revisit a vertex at any point, then we have discovered a cycle. Since G does not have any cycles, this is impossible. Thus, each vertex we visit must be a new vertex. Since there are only n vertices in G, after at most n steps we cannot discover any more new vertices. That means, the last vertex, the one we stopped at, must have degree 1. If there is an edge other than the one that we just used to arrive at the last vertex, then that edge would either lead to a previously visited vertex (creating a cycle) or to a new vertex (which implies that G has more than n vertices) — either of which leads to a contradiction.

Now, remove the degree 1 vertex we found from G. The resulting graph is still a connected graph without any cycles, which by the argument above, must have a degree 1 vertex. We keep removing the degree 1 vertices until all we have is a single vertex and no edges. Each time we removed a degree 1 vertex, we reduce the number vertices by 1 and the number of edges by 1. Let m be the number of degree 1 vertices we removed in the process. Then, e must equal m and n must equal m + 1, since the removal process took away all of the edges and all but one of the vertices. Thus, e = n − 1.

(II ⇒ III)   We prove this indirectly. That is, we show that ¬III ⇒ ¬II. In other words, we will show that if there are two paths in G between some pair of vertices u and v, then e ≠ n - 1. So, suppose that there are two paths π₁ and π₂ from u to v:

As we proved in Example 2 of the Uniqueness Proofs section, there must be a cycle from x back to itself going through y, w and z. Here x, y, z and w are vertices on π₁ and π₂ shown in the figure above. We note that every vertex in this cycle has degree at least 2.

As before, we keep removing degree 1 vertices from G until there are no more degree 1 vertices. Now, we are not allowed to assume that G does not have any cycles, since in proving II ⇒ III we are not allowed to assume I. So, it might be the case that G does not have degree 1 vertices, but that's OK. This argument takes a different route. Suppose that m vertices are removed when we remove degree 1 vertices. (It is possible that m = 0.) Let G' be the resulting graph.

Consider the cycle from x to itself that we found above — the one going through y, w and z. None of the vertices in this cycle was removed. Why is this? Suppose some vertex a is the first vertex from the cycle to be removed. But, if a was first, then a was still connected to two vertices in the cycle. That means, a still had degree 2 and should not have been removed — a contradiction. Thus, the process of removing degree 1 vertices will leave at least 4 vertices in G'.

Let n' and e' be respectively the number of vertices and edges in G'. Every vertex in G' has degree at least 2, so we have:

2 e'   =	∑	degree(v)
	v ∈ G'

and

∑	degree(v)   ≥   2 n'.
v ∈ G'

We know summing the degrees always gives us twice the number of edges. That accounts for the first equation. The second comes from every vertex in G' having degree at least 2. Thus, 2 e' ≥ 2 n' which implies that e' ≥ n'. Finally, in terms of the number of vertices n and the number of edges e in the original graph G, we have e = e' + m and n = n' + m because m degree 1 vertices were removed from G to form G'. Thus,

e = e' + m ≥ n' + m = n.

Therefore, e is at least n (thus cannot be equal to n − 1) and we have shown that ¬III ⇒ ¬II.

(III ⇒ I)   We prove this indirectly and and show that if G has a cycle then the path between some pair of vertices u and v is not unique. So, suppose that G has a cycle x₁, x₂, x₃, ..., x_k, x₁. Then, let u = x₁ and v = x₂. The edge between x₁ and x₂ constitutes a path from u to v. Furthermore, the reversal of the portion of the cycle from x₂ to x_k back to x₁ is also a path from u to v. Thus, there are two paths in G from u to v.

QED

x ∈ A ⇒ x ∈ B

x ∈ B ⇒ x ∈ A.

So, let's prove the following claim.

Claim: For all sets A, B and C,

( A ∪ B ) ∩ ( A ∪ C) = A ∪ ( (B ∩ C) − A ).

Proof:

(LHS ⊆ RHS)   Suppose that x ∈ ( A ∪ B ) ∩ ( A ∪ C). There are two cases to consider.

Case 1: x ∈ A.
If x ∈ A, then x ∈ A ∪ ( (B ∩ C) − A ).

Case 2: x ∉ A.
Since x ∈ ( A ∪ B ) ∩ ( A ∪ C), we know that x ∈ A ∪ B and x ∈ A ∪ C. But x ∉ A and x ∈ A ∪ B, implies that x ∈ B. Similarly, x ∉ A and x ∈ A ∪ C, implies that x ∈ C. Thus, x ∈ B ∩ C.

Furthermore, x ∉ A, so x ∈ (B ∩ C) − A. Therefore, x ∈ A ∪ ( (B ∩ C) − A ).

(LHS ⊆ RHS)   Suppose that x ∈ A ∪ ( (B ∩ C) − A ). Then, either x ∈ A or x ∈ ( (B ∩ C) − A ). Let's consider these case separately.

Case 1: x ∈ A.
If x ∈ A, then x ∈ A ∪ B. Similarly, x ∈ A ∪ C. Thus, x ∈ ( A ∪ B ) ∩ ( A ∪ C).

Case 2: x ∈ (B ∩ C) − A.
If x ∈ (B ∩ C) − A, then If x ∈ B ∩ C and x ∉ A. Since x ∈ B ∩ C, we know that x ∈ B and x ∈ C. Thus, x ∈ A ∪ B and x ∈ A ∪ C. Therefore, x ∈ ( A ∪ B ) ∩ ( A ∪ C).