Metamath Proof Explorer


Theorem usgrvd0nedg

Description: If a vertex in a simple graph has degree 0, the vertex is not adjacent to another vertex via an edge. (Contributed by Alexander van der Vekens, 20-Dec-2017) (Revised by AV, 16-Dec-2020) (Proof shortened by AV, 23-Dec-2020)

Ref Expression
Hypotheses vtxdusgradjvtx.v V = Vtx G
vtxdusgradjvtx.e E = Edg G
Assertion usgrvd0nedg G USGraph U V VtxDeg G U = 0 ¬ v V U v E

Proof

Step Hyp Ref Expression
1 vtxdusgradjvtx.v V = Vtx G
2 vtxdusgradjvtx.e E = Edg G
3 1 2 vtxdusgradjvtx G USGraph U V VtxDeg G U = v V | U v E
4 3 eqeq1d G USGraph U V VtxDeg G U = 0 v V | U v E = 0
5 1 fvexi V V
6 5 rabex v V | U v E V
7 hasheq0 v V | U v E V v V | U v E = 0 v V | U v E =
8 6 7 ax-mp v V | U v E = 0 v V | U v E =
9 rabeq0 v V | U v E = v V ¬ U v E
10 ralnex v V ¬ U v E ¬ v V U v E
11 10 biimpi v V ¬ U v E ¬ v V U v E
12 11 a1i G USGraph U V v V ¬ U v E ¬ v V U v E
13 9 12 syl5bi G USGraph U V v V | U v E = ¬ v V U v E
14 8 13 syl5bi G USGraph U V v V | U v E = 0 ¬ v V U v E
15 4 14 sylbid G USGraph U V VtxDeg G U = 0 ¬ v V U v E