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 FPathsGPFTrailsGPFunP1..^F-1P0FP1..^F=

Proof

Step Hyp Ref Expression
1 pthsfval PathsG=fp|fTrailsGpFunp1..^f-1p0fp1..^f=
2 3anass fTrailsGpFunp1..^f-1p0fp1..^f=fTrailsGpFunp1..^f-1p0fp1..^f=
3 2 opabbii fp|fTrailsGpFunp1..^f-1p0fp1..^f==fp|fTrailsGpFunp1..^f-1p0fp1..^f=
4 1 3 eqtri PathsG=fp|fTrailsGpFunp1..^f-1p0fp1..^f=
5 simpr f=Fp=Pp=P
6 fveq2 f=Ff=F
7 6 oveq2d f=F1..^f=1..^F
8 7 adantr f=Fp=P1..^f=1..^F
9 5 8 reseq12d f=Fp=Pp1..^f=P1..^F
10 9 cnveqd f=Fp=Pp1..^f-1=P1..^F-1
11 10 funeqd f=Fp=PFunp1..^f-1FunP1..^F-1
12 6 preq2d f=F0f=0F
13 12 adantr f=Fp=P0f=0F
14 5 13 imaeq12d f=Fp=Pp0f=P0F
15 5 8 imaeq12d f=Fp=Pp1..^f=P1..^F
16 14 15 ineq12d f=Fp=Pp0fp1..^f=P0FP1..^F
17 16 eqeq1d f=Fp=Pp0fp1..^f=P0FP1..^F=
18 11 17 anbi12d f=Fp=PFunp1..^f-1p0fp1..^f=FunP1..^F-1P0FP1..^F=
19 reltrls RelTrailsG
20 4 18 19 brfvopabrbr FPathsGPFTrailsGPFunP1..^F-1P0FP1..^F=
21 3anass FTrailsGPFunP1..^F-1P0FP1..^F=FTrailsGPFunP1..^F-1P0FP1..^F=
22 20 21 bitr4i FPathsGPFTrailsGPFunP1..^F-1P0FP1..^F=