Metamath Proof Explorer

Theorem umgrnloop

Description: In a multigraph, there is no loop, i.e. no edge connecting a vertex with itself. (Contributed by Alexander van der Vekens, 19-Aug-2017) (Revised by AV, 11-Dec-2020)

		Ref	Expression
	Hypothesis	umgrnloopv.e	$⊢ E = iEdg (G)$
	Assertion	umgrnloop	$⊢ G \in UMGraph \to (\exists x \in dom E E (x) = \{M, N\} \to M \neq N)$

Proof

Step	Hyp	Ref	Expression
1		umgrnloopv.e	$⊢ E = iEdg (G)$
2		eqid	$⊢ Vtx (G) = Vtx (G)$
3	1 2	umgredgprv	$⊢ (G \in UMGraph \land x \in dom E) \to (E (x) = \{M, N\} \to (M \in Vtx (G) \land N \in Vtx (G)))$
4	3	imp	$⊢ ((G \in UMGraph \land x \in dom E) \land E (x) = \{M, N\}) \to (M \in Vtx (G) \land N \in Vtx (G))$
5	1	umgrnloopv	$⊢ (G \in UMGraph \land M \in Vtx (G)) \to (E (x) = \{M, N\} \to M \neq N)$
6	5	ex	$⊢ G \in UMGraph \to (M \in Vtx (G) \to (E (x) = \{M, N\} \to M \neq N))$
7	6	com23	$⊢ G \in UMGraph \to (E (x) = \{M, N\} \to (M \in Vtx (G) \to M \neq N))$
8	7	adantr	$⊢ (G \in UMGraph \land x \in dom E) \to (E (x) = \{M, N\} \to (M \in Vtx (G) \to M \neq N))$
9	8	imp	$⊢ ((G \in UMGraph \land x \in dom E) \land E (x) = \{M, N\}) \to (M \in Vtx (G) \to M \neq N)$
10	9	com12	$⊢ M \in Vtx (G) \to (((G \in UMGraph \land x \in dom E) \land E (x) = \{M, N\}) \to M \neq N)$
11	10	adantr	$⊢ (M \in Vtx (G) \land N \in Vtx (G)) \to (((G \in UMGraph \land x \in dom E) \land E (x) = \{M, N\}) \to M \neq N)$
12	4 11	mpcom	$⊢ ((G \in UMGraph \land x \in dom E) \land E (x) = \{M, N\}) \to M \neq N$
13	12	rexlimdva2	$⊢ G \in UMGraph \to (\exists x \in dom E E (x) = \{M, N\} \to M \neq N)$