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