Theorem vtxdg0v 26369

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.)

Hypothesis

Ref	Expression
vtxdgf.v	⊢ 𝑉 = (Vtx‘𝐺)

Assertion

Ref	Expression
vtxdg0v	⊢ ((𝐺 = ∅ ∧ 𝑈 ∈ 𝑉) → ((VtxDeg‘𝐺)‘𝑈) = 0)

Proof of Theorem vtxdg0v

Step	Hyp	Ref	Expression
1		vtxdgf.v	. . . . 5 ⊢ 𝑉 = (Vtx‘𝐺)
2	1	eleq2i 2693	. . . 4 ⊢ (𝑈 ∈ 𝑉 ↔ 𝑈 ∈ (Vtx‘𝐺))
3		fveq2 6191	. . . . . 6 ⊢ (𝐺 = ∅ → (Vtx‘𝐺) = (Vtx‘∅))
4		vtxval0 25931	. . . . . 6 ⊢ (Vtx‘∅) = ∅
5	3, 4	syl6eq 2672	. . . . 5 ⊢ (𝐺 = ∅ → (Vtx‘𝐺) = ∅)
6	5	eleq2d 2687	. . . 4 ⊢ (𝐺 = ∅ → (𝑈 ∈ (Vtx‘𝐺) ↔ 𝑈 ∈ ∅))
7	2, 6	syl5bb 272	. . 3 ⊢ (𝐺 = ∅ → (𝑈 ∈ 𝑉 ↔ 𝑈 ∈ ∅))
8		noel 3919	. . . 4 ⊢ ¬ 𝑈 ∈ ∅
9	8	pm2.21i 116	. . 3 ⊢ (𝑈 ∈ ∅ → ((VtxDeg‘𝐺)‘𝑈) = 0)
10	7, 9	syl6bi 243	. 2 ⊢ (𝐺 = ∅ → (𝑈 ∈ 𝑉 → ((VtxDeg‘𝐺)‘𝑈) = 0))
11	10	imp 445	1 ⊢ ((𝐺 = ∅ ∧ 𝑈 ∈ 𝑉) → ((VtxDeg‘𝐺)‘𝑈) = 0)

Syntax hints:   → wi 4   ∧ wa 384   = wceq 1483   ∈ wcel 1990  ∅c0 3915  ‘cfv 5888  0cc0 9936  Vtxcvtx 25874  VtxDegcvtxdg 26361

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-8 1992  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781  ax-nul 4789  ax-pow 4843  ax-pr 4906

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-eu 2474  df-mo 2475  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-ral 2917  df-rex 2918  df-rab 2921  df-v 3202  df-sbc 3436  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184  df-uni 4437  df-br 4654  df-opab 4713  df-mpt 4730  df-id 5024  df-xp 5120  df-rel 5121  df-cnv 5122  df-co 5123  df-dm 5124  df-iota 5851  df-fun 5890  df-fv 5896  df-slot 15861  df-base 15863  df-vtx 25876

This theorem is referenced by: (None)