Metamath Proof Explorer


Theorem cocnvf1o

Description: Composing with the inverse of a bijection. (Contributed by Thierry Arnoux, 15-Jan-2026)

Ref Expression
Hypotheses cocnvf1o.1 φ F : A B
cocnvf1o.2 φ G : A B
cocnvf1o.3 φ H : A 1-1 onto A
Assertion cocnvf1o φ F = G H G = F H -1

Proof

Step Hyp Ref Expression
1 cocnvf1o.1 φ F : A B
2 cocnvf1o.2 φ G : A B
3 cocnvf1o.3 φ H : A 1-1 onto A
4 simpr φ F = G H F = G H
5 4 coeq1d φ F = G H F H -1 = G H H -1
6 coass G H H -1 = G H H -1
7 5 6 eqtrdi φ F = G H F H -1 = G H H -1
8 f1ococnv2 H : A 1-1 onto A H H -1 = I A
9 3 8 syl φ H H -1 = I A
10 9 coeq2d φ G H H -1 = G I A
11 fcoi1 G : A B G I A = G
12 2 11 syl φ G I A = G
13 10 12 eqtrd φ G H H -1 = G
14 13 adantr φ F = G H G H H -1 = G
15 7 14 eqtr2d φ F = G H G = F H -1
16 simpr φ G = F H -1 G = F H -1
17 16 coeq1d φ G = F H -1 G H = F H -1 H
18 coass F H -1 H = F H -1 H
19 17 18 eqtrdi φ G = F H -1 G H = F H -1 H
20 f1ococnv1 H : A 1-1 onto A H -1 H = I A
21 3 20 syl φ H -1 H = I A
22 21 coeq2d φ F H -1 H = F I A
23 fcoi1 F : A B F I A = F
24 1 23 syl φ F I A = F
25 22 24 eqtrd φ F H -1 H = F
26 25 adantr φ G = F H -1 F H -1 H = F
27 19 26 eqtr2d φ G = F H -1 F = G H
28 15 27 impbida φ F = G H G = F H -1