Metamath Proof Explorer


Theorem ispth

Description: Conditions for a pair of classes/functions to be a path (in an undirected graph). (Contributed by Alexander van der Vekens, 21-Oct-2017) (Revised by AV, 9-Jan-2021) (Revised by AV, 29-Oct-2021)

Ref Expression
Assertion ispth F Paths G P F Trails G P Fun P 1 ..^ F -1 P 0 F P 1 ..^ F =

Proof

Step Hyp Ref Expression
1 pthsfval Paths G = f p | f Trails G p Fun p 1 ..^ f -1 p 0 f p 1 ..^ f =
2 3anass f Trails G p Fun p 1 ..^ f -1 p 0 f p 1 ..^ f = f Trails G p Fun p 1 ..^ f -1 p 0 f p 1 ..^ f =
3 2 opabbii f p | f Trails G p Fun p 1 ..^ f -1 p 0 f p 1 ..^ f = = f p | f Trails G p Fun p 1 ..^ f -1 p 0 f p 1 ..^ f =
4 1 3 eqtri Paths G = f p | f Trails G p Fun p 1 ..^ f -1 p 0 f p 1 ..^ f =
5 simpr f = F p = P p = P
6 fveq2 f = F f = F
7 6 oveq2d f = F 1 ..^ f = 1 ..^ F
8 7 adantr f = F p = P 1 ..^ f = 1 ..^ F
9 5 8 reseq12d f = F p = P p 1 ..^ f = P 1 ..^ F
10 9 cnveqd f = F p = P p 1 ..^ f -1 = P 1 ..^ F -1
11 10 funeqd f = F p = P Fun p 1 ..^ f -1 Fun P 1 ..^ F -1
12 6 preq2d f = F 0 f = 0 F
13 12 adantr f = F p = P 0 f = 0 F
14 5 13 imaeq12d f = F p = P p 0 f = P 0 F
15 5 8 imaeq12d f = F p = P p 1 ..^ f = P 1 ..^ F
16 14 15 ineq12d f = F p = P p 0 f p 1 ..^ f = P 0 F P 1 ..^ F
17 16 eqeq1d f = F p = P p 0 f p 1 ..^ f = P 0 F P 1 ..^ F =
18 11 17 anbi12d f = F p = P Fun p 1 ..^ f -1 p 0 f p 1 ..^ f = Fun P 1 ..^ F -1 P 0 F P 1 ..^ F =
19 reltrls Rel Trails G
20 4 18 19 brfvopabrbr F Paths G P F Trails G P Fun P 1 ..^ F -1 P 0 F P 1 ..^ F =
21 3anass F Trails G P Fun P 1 ..^ F -1 P 0 F P 1 ..^ F = F Trails G P Fun P 1 ..^ F -1 P 0 F P 1 ..^ F =
22 20 21 bitr4i F Paths G P F Trails G P Fun P 1 ..^ F -1 P 0 F P 1 ..^ F =