Metamath Proof Explorer

Theorem upgrwlkedg

Description: The edges of a walk in a pseudograph join exactly the two corresponding adjacent vertices in the walk. (Contributed by AV, 27-Feb-2021)

		Ref	Expression
	Hypothesis	upgrwlkedg.i	$⊢ I = iEdg (G)$
	Assertion	upgrwlkedg	$⊢ (G \in UPGraph \land F Walks (G) P) \to \forall k \in (0 ..^\|F\|) I (F (k)) = \{P (k), P (k + 1)\}$

Proof

Step	Hyp	Ref	Expression
1		upgrwlkedg.i	$⊢ I = iEdg (G)$
2		eqid	$⊢ Vtx (G) = Vtx (G)$
3	2 1	upgriswlk	$⊢ G \in UPGraph \to (F Walks (G) P \leftrightarrow (F \in Word dom I \land P : (0 \dots \|F\|) ⟶ Vtx (G) \land \forall k \in (0 ..^\|F\|) I (F (k)) = \{P (k), P (k + 1)\}))$
4		simp3	$⊢ (F \in Word dom I \land P : (0 \dots \|F\|) ⟶ Vtx (G) \land \forall k \in (0 ..^\|F\|) I (F (k)) = \{P (k), P (k + 1)\}) \to \forall k \in (0 ..^\|F\|) I (F (k)) = \{P (k), P (k + 1)\}$
5	3 4	syl6bi	$⊢ G \in UPGraph \to (F Walks (G) P \to \forall k \in (0 ..^\|F\|) I (F (k)) = \{P (k), P (k + 1)\})$
6	5	imp	$⊢ (G \in UPGraph \land F Walks (G) P) \to \forall k \in (0 ..^\|F\|) I (F (k)) = \{P (k), P (k + 1)\}$