Metamath Proof Explorer


Theorem eupth0

Description: There is an Eulerian path on an empty graph, i.e. a graph with at least one vertex, but without an edge. (Contributed by Mario Carneiro, 7-Apr-2015) (Revised by AV, 5-Mar-2021) (Proof shortened by AV, 30-Oct-2021)

Ref Expression
Hypotheses eupth0.v V=VtxG
eupth0.i I=iEdgG
Assertion eupth0 AVI=EulerPathsG0A

Proof

Step Hyp Ref Expression
1 eupth0.v V=VtxG
2 eupth0.i I=iEdgG
3 eqidd AV0A=0A
4 1 is0wlk 0A=0AAVWalksG0A
5 3 4 mpancom AVWalksG0A
6 f1o0 :1-1 onto
7 eqidd I==
8 hash0 =0
9 8 oveq2i 0..^=0..^0
10 fzo0 0..^0=
11 9 10 eqtri 0..^=
12 11 a1i I=0..^=
13 dmeq I=domI=dom
14 dm0 dom=
15 13 14 eqtrdi I=domI=
16 7 12 15 f1oeq123d I=:0..^1-1 ontodomI:1-1 onto
17 6 16 mpbiri I=:0..^1-1 ontodomI
18 5 17 anim12i AVI=WalksG0A:0..^1-1 ontodomI
19 2 iseupthf1o EulerPathsG0AWalksG0A:0..^1-1 ontodomI
20 18 19 sylibr AVI=EulerPathsG0A