Metamath Proof Explorer


Theorem nbusgredgeu0

Description: For each neighbor of a vertex there is exactly one edge between the vertex and its neighbor in a simple graph. (Contributed by Alexander van der Vekens, 17-Dec-2017) (Revised by AV, 27-Oct-2020) (Proof shortened by AV, 13-Feb-2022)

Ref Expression
Hypotheses nbusgrf1o1.v V = Vtx G
nbusgrf1o1.e E = Edg G
nbusgrf1o1.n N = G NeighbVtx U
nbusgrf1o1.i I = e E | U e
Assertion nbusgredgeu0 G USGraph U V M N ∃! i I i = U M

Proof

Step Hyp Ref Expression
1 nbusgrf1o1.v V = Vtx G
2 nbusgrf1o1.e E = Edg G
3 nbusgrf1o1.n N = G NeighbVtx U
4 nbusgrf1o1.i I = e E | U e
5 simpll G USGraph U V M N G USGraph
6 3 eleq2i M N M G NeighbVtx U
7 nbgrsym M G NeighbVtx U U G NeighbVtx M
8 7 a1i G USGraph U V M G NeighbVtx U U G NeighbVtx M
9 8 biimpd G USGraph U V M G NeighbVtx U U G NeighbVtx M
10 6 9 syl5bi G USGraph U V M N U G NeighbVtx M
11 10 imp G USGraph U V M N U G NeighbVtx M
12 2 nbusgredgeu G USGraph U G NeighbVtx M ∃! i E i = U M
13 5 11 12 syl2anc G USGraph U V M N ∃! i E i = U M
14 df-reu ∃! i E i = U M ∃! i i E i = U M
15 13 14 sylib G USGraph U V M N ∃! i i E i = U M
16 anass i E U i i = U M i E U i i = U M
17 prid1g U V U U M
18 17 ad2antlr G USGraph U V M N U U M
19 eleq2 i = U M U i U U M
20 18 19 syl5ibrcom G USGraph U V M N i = U M U i
21 20 pm4.71rd G USGraph U V M N i = U M U i i = U M
22 21 bicomd G USGraph U V M N U i i = U M i = U M
23 22 anbi2d G USGraph U V M N i E U i i = U M i E i = U M
24 16 23 syl5bb G USGraph U V M N i E U i i = U M i E i = U M
25 24 eubidv G USGraph U V M N ∃! i i E U i i = U M ∃! i i E i = U M
26 15 25 mpbird G USGraph U V M N ∃! i i E U i i = U M
27 df-reu ∃! i I i = U M ∃! i i I i = U M
28 eleq2 e = i U e U i
29 28 4 elrab2 i I i E U i
30 29 anbi1i i I i = U M i E U i i = U M
31 30 eubii ∃! i i I i = U M ∃! i i E U i i = U M
32 27 31 bitri ∃! i I i = U M ∃! i i E U i i = U M
33 26 32 sylibr G USGraph U V M N ∃! i I i = U M