Metamath Proof Explorer


Theorem snstrvtxval

Description: The set of vertices of a graph without edges represented as an extensible structure with vertices as base set and no indexed edges. See vtxvalsnop for the (degenerate) case where V = ( Basendx ) . (Contributed by AV, 23-Sep-2020)

Ref Expression
Hypotheses snstrvtxval.v 𝑉 ∈ V
snstrvtxval.g 𝐺 = { ⟨ ( Base ‘ ndx ) , 𝑉 ⟩ }
Assertion snstrvtxval ( 𝑉 ≠ ( Base ‘ ndx ) → ( Vtx ‘ 𝐺 ) = 𝑉 )

Proof

Step Hyp Ref Expression
1 snstrvtxval.v 𝑉 ∈ V
2 snstrvtxval.g 𝐺 = { ⟨ ( Base ‘ ndx ) , 𝑉 ⟩ }
3 necom ( 𝑉 ≠ ( Base ‘ ndx ) ↔ ( Base ‘ ndx ) ≠ 𝑉 )
4 fvex ( Base ‘ ndx ) ∈ V
5 4 1 2 funsndifnop ( ( Base ‘ ndx ) ≠ 𝑉 → ¬ 𝐺 ∈ ( V × V ) )
6 3 5 sylbi ( 𝑉 ≠ ( Base ‘ ndx ) → ¬ 𝐺 ∈ ( V × V ) )
7 6 iffalsed ( 𝑉 ≠ ( Base ‘ ndx ) → if ( 𝐺 ∈ ( V × V ) , ( 1st𝐺 ) , ( Base ‘ 𝐺 ) ) = ( Base ‘ 𝐺 ) )
8 vtxval ( Vtx ‘ 𝐺 ) = if ( 𝐺 ∈ ( V × V ) , ( 1st𝐺 ) , ( Base ‘ 𝐺 ) )
9 8 a1i ( 𝑉 ≠ ( Base ‘ ndx ) → ( Vtx ‘ 𝐺 ) = if ( 𝐺 ∈ ( V × V ) , ( 1st𝐺 ) , ( Base ‘ 𝐺 ) ) )
10 2 1strbas ( 𝑉 ∈ V → 𝑉 = ( Base ‘ 𝐺 ) )
11 1 10 mp1i ( 𝑉 ≠ ( Base ‘ ndx ) → 𝑉 = ( Base ‘ 𝐺 ) )
12 7 9 11 3eqtr4d ( 𝑉 ≠ ( Base ‘ ndx ) → ( Vtx ‘ 𝐺 ) = 𝑉 )