Metamath Proof Explorer


Theorem usgrexmpl1vtx

Description: The vertices 0 , 1 , 2 , 3 , 4 , 5 of the graph G = <. V , E >. . (Contributed by AV, 3-Aug-2025)

Ref Expression
Hypotheses usgrexmpl1.v V = 0 5
usgrexmpl1.e E = ⟨“ 0 1 0 2 1 2 0 3 3 4 3 5 4 5 ”⟩
usgrexmpl1.g G = V E
Assertion usgrexmpl1vtx Vtx G = 0 1 2 3 4 5

Proof

Step Hyp Ref Expression
1 usgrexmpl1.v V = 0 5
2 usgrexmpl1.e E = ⟨“ 0 1 0 2 1 2 0 3 3 4 3 5 4 5 ”⟩
3 usgrexmpl1.g G = V E
4 3 fveq2i Vtx G = Vtx V E
5 1 ovexi V V
6 s7cli ⟨“ 0 1 0 2 1 2 0 3 3 4 3 5 4 5 ”⟩ Word V
7 2 6 eqeltri E Word V
8 opvtxfv V V E Word V Vtx V E = V
9 5 7 8 mp2an Vtx V E = V
10 4 9 eqtri Vtx G = V
11 fz0to5un2tp 0 5 = 0 1 2 3 4 5
12 10 1 11 3eqtri Vtx G = 0 1 2 3 4 5