Metamath Proof Explorer


Theorem p1evtxdeqlem

Description: Lemma for p1evtxdeq and p1evtxdp1 . (Contributed by AV, 3-Mar-2021)

Ref Expression
Hypotheses p1evtxdeq.v ⊢ V = Vtx ⁡ G
p1evtxdeq.i ⊢ I = iEdg ⁡ G
p1evtxdeq.f ⊢ φ → Fun ⁡ I
p1evtxdeq.fv ⊢ φ → Vtx ⁡ F = V
p1evtxdeq.fi ⊢ φ → iEdg ⁡ F = I ∪ K E
p1evtxdeq.k ⊢ φ → K ∈ X
p1evtxdeq.d ⊢ φ → K ∉ dom ⁡ I
p1evtxdeq.u ⊢ φ → U ∈ V
p1evtxdeq.e ⊢ φ → E ∈ Y
Assertion p1evtxdeqlem ⊢ φ → VtxDeg ⁡ F ⁡ U = VtxDeg ⁡ G ⁡ U + 𝑒 VtxDeg ⁡ V K E ⁡ U

Proof

Step Hyp Ref Expression
1 p1evtxdeq.v ⊢ V = Vtx ⁡ G
2 p1evtxdeq.i ⊢ I = iEdg ⁡ G
3 p1evtxdeq.f ⊢ φ → Fun ⁡ I
4 p1evtxdeq.fv ⊢ φ → Vtx ⁡ F = V
5 p1evtxdeq.fi ⊢ φ → iEdg ⁡ F = I ∪ K E
6 p1evtxdeq.k ⊢ φ → K ∈ X
7 p1evtxdeq.d ⊢ φ → K ∉ dom ⁡ I
8 p1evtxdeq.u ⊢ φ → U ∈ V
9 p1evtxdeq.e ⊢ φ → E ∈ Y
10 1 fvexi ⊢ V ∈ V
11 snex ⊢ K E ∈ V
12 10 11 pm3.2i ⊢ V ∈ V ∧ K E ∈ V
13 opiedgfv ⊢ V ∈ V ∧ K E ∈ V → iEdg ⁡ V K E = K E
14 13 eqcomd ⊢ V ∈ V ∧ K E ∈ V → K E = iEdg ⁡ V K E
15 12 14 ax-mp ⊢ K E = iEdg ⁡ V K E
16 opvtxfv ⊢ V ∈ V ∧ K E ∈ V → Vtx ⁡ V K E = V
17 12 16 mp1i ⊢ φ → Vtx ⁡ V K E = V
18 dmsnopg ⊢ E ∈ Y → dom ⁡ K E = K
19 9 18 syl ⊢ φ → dom ⁡ K E = K
20 19 ineq2d ⊢ φ → dom ⁡ I ∩ dom ⁡ K E = dom ⁡ I ∩ K
21 df-nel ⊢ K ∉ dom ⁡ I ↔ ¬ K ∈ dom ⁡ I
22 7 21 sylib ⊢ φ → ¬ K ∈ dom ⁡ I
23 disjsn ⊢ dom ⁡ I ∩ K = ∅ ↔ ¬ K ∈ dom ⁡ I
24 22 23 sylibr ⊢ φ → dom ⁡ I ∩ K = ∅
25 20 24 eqtrd ⊢ φ → dom ⁡ I ∩ dom ⁡ K E = ∅
26 funsng ⊢ K ∈ X ∧ E ∈ Y → Fun ⁡ K E
27 6 9 26 syl2anc ⊢ φ → Fun ⁡ K E
28 2 15 1 17 4 25 3 27 8 5 vtxdun ⊢ φ → VtxDeg ⁡ F ⁡ U = VtxDeg ⁡ G ⁡ U + 𝑒 VtxDeg ⁡ V K E ⁡ U