Metamath Proof Explorer


Theorem usgr1v0e

Description: The size of a (finite) simple graph with 1 vertex is 0. (Contributed by Alexander van der Vekens, 5-Jan-2018) (Revised by AV, 22-Oct-2020)

Ref Expression
Hypotheses fusgredgfi.v V = Vtx G
fusgredgfi.e E = Edg G
Assertion usgr1v0e G USGraph V = 1 E = 0

Proof

Step Hyp Ref Expression
1 fusgredgfi.v V = Vtx G
2 fusgredgfi.e E = Edg G
3 simpl G USGraph V = v G USGraph
4 vex v V
5 1 eqeq1i V = v Vtx G = v
6 5 biimpi V = v Vtx G = v
7 6 adantl G USGraph V = v Vtx G = v
8 usgr1vr v V Vtx G = v G USGraph iEdg G =
9 4 7 8 sylancr G USGraph V = v G USGraph iEdg G =
10 3 9 mpd G USGraph V = v iEdg G =
11 2 eqeq1i E = Edg G =
12 usgruhgr G USGraph G UHGraph
13 uhgriedg0edg0 G UHGraph Edg G = iEdg G =
14 12 13 syl G USGraph Edg G = iEdg G =
15 14 adantr G USGraph V = v Edg G = iEdg G =
16 11 15 syl5bb G USGraph V = v E = iEdg G =
17 10 16 mpbird G USGraph V = v E =
18 17 ex G USGraph V = v E =
19 18 exlimdv G USGraph v V = v E =
20 1 fvexi V V
21 hash1snb V V V = 1 v V = v
22 20 21 mp1i G USGraph V = 1 v V = v
23 2 fvexi E V
24 hasheq0 E V E = 0 E =
25 23 24 mp1i G USGraph E = 0 E =
26 19 22 25 3imtr4d G USGraph V = 1 E = 0
27 26 imp G USGraph V = 1 E = 0