Metamath Proof Explorer


Theorem updjudhcoinlf

Description: The composition of the mapping of an element of the disjoint union to the value of the corresponding function and the left injection equals the first function. (Contributed by AV, 27-Jun-2022)

Ref Expression
Hypotheses updjud.f φF:AC
updjud.g φG:BC
updjudhf.h H=xA⊔︀Bif1stx=F2ndxG2ndx
Assertion updjudhcoinlf φHinlA=F

Proof

Step Hyp Ref Expression
1 updjud.f φF:AC
2 updjud.g φG:BC
3 updjudhf.h H=xA⊔︀Bif1stx=F2ndxG2ndx
4 1 2 3 updjudhf φH:A⊔︀BC
5 4 ffnd φHFnA⊔︀B
6 inlresf inlA:AA⊔︀B
7 ffn inlA:AA⊔︀BinlAFnA
8 6 7 mp1i φinlAFnA
9 frn inlA:AA⊔︀BraninlAA⊔︀B
10 6 9 mp1i φraninlAA⊔︀B
11 fnco HFnA⊔︀BinlAFnAraninlAA⊔︀BHinlAFnA
12 5 8 10 11 syl3anc φHinlAFnA
13 1 ffnd φFFnA
14 fvco2 inlAFnAaAHinlAa=HinlAa
15 8 14 sylan φaAHinlAa=HinlAa
16 fvres aAinlAa=inla
17 16 adantl φaAinlAa=inla
18 17 fveq2d φaAHinlAa=Hinla
19 fveqeq2 x=inla1stx=1stinla=
20 2fveq3 x=inlaF2ndx=F2ndinla
21 2fveq3 x=inlaG2ndx=G2ndinla
22 19 20 21 ifbieq12d x=inlaif1stx=F2ndxG2ndx=if1stinla=F2ndinlaG2ndinla
23 22 adantl φaAx=inlaif1stx=F2ndxG2ndx=if1stinla=F2ndinlaG2ndinla
24 1stinl aA1stinla=
25 24 adantl φaA1stinla=
26 25 adantr φaAx=inla1stinla=
27 26 iftrued φaAx=inlaif1stinla=F2ndinlaG2ndinla=F2ndinla
28 23 27 eqtrd φaAx=inlaif1stx=F2ndxG2ndx=F2ndinla
29 djulcl aAinlaA⊔︀B
30 29 adantl φaAinlaA⊔︀B
31 1 adantr φaAF:AC
32 2ndinl aA2ndinla=a
33 32 adantl φaA2ndinla=a
34 simpr φaAaA
35 33 34 eqeltrd φaA2ndinlaA
36 31 35 ffvelcdmd φaAF2ndinlaC
37 3 28 30 36 fvmptd2 φaAHinla=F2ndinla
38 18 37 eqtrd φaAHinlAa=F2ndinla
39 33 fveq2d φaAF2ndinla=Fa
40 15 38 39 3eqtrd φaAHinlAa=Fa
41 12 13 40 eqfnfvd φHinlA=F