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