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 = Vtx G
eupth0.i I = iEdg G
Assertion eupth0 A V I = EulerPaths G 0 A

Proof

Step Hyp Ref Expression
1 eupth0.v V = Vtx G
2 eupth0.i I = iEdg G
3 eqidd A V 0 A = 0 A
4 1 is0wlk 0 A = 0 A A V Walks G 0 A
5 3 4 mpancom A V Walks G 0 A
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 = dom I = dom
14 dm0 dom =
15 13 14 eqtrdi I = dom I =
16 7 12 15 f1oeq123d I = : 0 ..^ 1-1 onto dom I : 1-1 onto
17 6 16 mpbiri I = : 0 ..^ 1-1 onto dom I
18 5 17 anim12i A V I = Walks G 0 A : 0 ..^ 1-1 onto dom I
19 2 iseupthf1o EulerPaths G 0 A Walks G 0 A : 0 ..^ 1-1 onto dom I
20 18 19 sylibr A V I = EulerPaths G 0 A