Metamath Proof Explorer


Theorem vtxdg0v

Description: The degree of a vertex in the null graph is zero (or anything else), because there are no vertices. (Contributed by AV, 11-Dec-2020)

Ref Expression
Hypothesis vtxdgf.v 𝑉 = ( Vtx ‘ 𝐺 )
Assertion vtxdg0v ( ( 𝐺 = ∅ ∧ 𝑈𝑉 ) → ( ( VtxDeg ‘ 𝐺 ) ‘ 𝑈 ) = 0 )

Proof

Step Hyp Ref Expression
1 vtxdgf.v 𝑉 = ( Vtx ‘ 𝐺 )
2 1 eleq2i ( 𝑈𝑉𝑈 ∈ ( Vtx ‘ 𝐺 ) )
3 fveq2 ( 𝐺 = ∅ → ( Vtx ‘ 𝐺 ) = ( Vtx ‘ ∅ ) )
4 vtxval0 ( Vtx ‘ ∅ ) = ∅
5 3 4 eqtrdi ( 𝐺 = ∅ → ( Vtx ‘ 𝐺 ) = ∅ )
6 5 eleq2d ( 𝐺 = ∅ → ( 𝑈 ∈ ( Vtx ‘ 𝐺 ) ↔ 𝑈 ∈ ∅ ) )
7 2 6 syl5bb ( 𝐺 = ∅ → ( 𝑈𝑉𝑈 ∈ ∅ ) )
8 noel ¬ 𝑈 ∈ ∅
9 8 pm2.21i ( 𝑈 ∈ ∅ → ( ( VtxDeg ‘ 𝐺 ) ‘ 𝑈 ) = 0 )
10 7 9 syl6bi ( 𝐺 = ∅ → ( 𝑈𝑉 → ( ( VtxDeg ‘ 𝐺 ) ‘ 𝑈 ) = 0 ) )
11 10 imp ( ( 𝐺 = ∅ ∧ 𝑈𝑉 ) → ( ( VtxDeg ‘ 𝐺 ) ‘ 𝑈 ) = 0 )