Metamath Proof Explorer

Theorem usgr2edg1

Description: If a vertex is adjacent to two different vertices in a simple graph, there is not only one edge starting at this vertex. (Contributed by Alexander van der Vekens, 10-Dec-2017) (Revised by AV, 17-Oct-2020) (Proof shortened by AV, 8-Jun-2021)

	Ref	Expression
Hypotheses	usgrf1oedg.i	$⊢ I = iEdg (G)$
	usgrf1oedg.e	$⊢ E = Edg (G)$
Assertion	usgr2edg1	$⊢ ((G \in USGraph \land A \neq B) \land (\{N, A\} \in E \land \{B, N\} \in E)) \to \neg \exists! x \in dom I N \in I (x)$

Proof

Step	Hyp	Ref	Expression
1		usgrf1oedg.i	$⊢ I = iEdg (G)$
2		usgrf1oedg.e	$⊢ E = Edg (G)$
3		usgrumgr	$⊢ G \in USGraph \to G \in UMGraph$
4	1 2	umgr2edg1	$⊢ ((G \in UMGraph \land A \neq B) \land (\{N, A\} \in E \land \{B, N\} \in E)) \to \neg \exists! x \in dom I N \in I (x)$
5	3 4	sylanl1	$⊢ ((G \in USGraph \land A \neq B) \land (\{N, A\} \in E \land \{B, N\} \in E)) \to \neg \exists! x \in dom I N \in I (x)$