Metamath Proof Explorer


Theorem foeqcnvco

Description: Condition for function equality in terms of vanishing of the composition with the converse.EDITORIAL: Is there a relation-algebraic proof of this? (Contributed by Stefan O'Rear, 12-Feb-2015)

Ref Expression
Assertion foeqcnvco ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B → F = G ↔ F ∘ G -1 = I ↾ B

Proof

Step Hyp Ref Expression
1 fococnv2 ⊢ F : A ⟶ onto B → F ∘ F -1 = I ↾ B
2 cnveq ⊢ F = G → F -1 = G -1
3 2 coeq2d ⊢ F = G → F ∘ F -1 = F ∘ G -1
4 3 eqeq1d ⊢ F = G → F ∘ F -1 = I ↾ B ↔ F ∘ G -1 = I ↾ B
5 1 4 syl5ibcom ⊢ F : A ⟶ onto B → F = G → F ∘ G -1 = I ↾ B
6 5 adantr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B → F = G → F ∘ G -1 = I ↾ B
7 fofn ⊢ F : A ⟶ onto B → F Fn A
8 7 ad2antrr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B → F Fn A
9 fofn ⊢ G : A ⟶ onto B → G Fn A
10 9 ad2antlr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B → G Fn A
11 9 adantl ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B → G Fn A
12 fnopfv ⊢ G Fn A ∧ x ∈ A → x G ⁡ x ∈ G
13 11 12 sylan ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → x G ⁡ x ∈ G
14 fvex ⊢ G ⁡ x ∈ V
15 vex ⊢ x ∈ V
16 14 15 brcnv ⊢ G ⁡ x G -1 x ↔ x G G ⁡ x
17 df-br ⊢ x G G ⁡ x ↔ x G ⁡ x ∈ G
18 16 17 bitri ⊢ G ⁡ x G -1 x ↔ x G ⁡ x ∈ G
19 13 18 sylibr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → G ⁡ x G -1 x
20 7 adantr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B → F Fn A
21 fnopfv ⊢ F Fn A ∧ x ∈ A → x F ⁡ x ∈ F
22 20 21 sylan ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → x F ⁡ x ∈ F
23 df-br ⊢ x F F ⁡ x ↔ x F ⁡ x ∈ F
24 22 23 sylibr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → x F F ⁡ x
25 breq2 ⊢ y = x → G ⁡ x G -1 y ↔ G ⁡ x G -1 x
26 breq1 ⊢ y = x → y F F ⁡ x ↔ x F F ⁡ x
27 25 26 anbi12d ⊢ y = x → G ⁡ x G -1 y ∧ y F F ⁡ x ↔ G ⁡ x G -1 x ∧ x F F ⁡ x
28 15 27 spcev ⊢ G ⁡ x G -1 x ∧ x F F ⁡ x → ∃ y G ⁡ x G -1 y ∧ y F F ⁡ x
29 19 24 28 syl2anc ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → ∃ y G ⁡ x G -1 y ∧ y F F ⁡ x
30 fvex ⊢ F ⁡ x ∈ V
31 14 30 brco ⊢ G ⁡ x F ∘ G -1 F ⁡ x ↔ ∃ y G ⁡ x G -1 y ∧ y F F ⁡ x
32 29 31 sylibr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → G ⁡ x F ∘ G -1 F ⁡ x
33 32 adantlr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B ∧ x ∈ A → G ⁡ x F ∘ G -1 F ⁡ x
34 breq ⊢ F ∘ G -1 = I ↾ B → G ⁡ x F ∘ G -1 F ⁡ x ↔ G ⁡ x I ↾ B F ⁡ x
35 34 ad2antlr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B ∧ x ∈ A → G ⁡ x F ∘ G -1 F ⁡ x ↔ G ⁡ x I ↾ B F ⁡ x
36 33 35 mpbid ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B ∧ x ∈ A → G ⁡ x I ↾ B F ⁡ x
37 fof ⊢ G : A ⟶ onto B → G : A ⟶ B
38 37 adantl ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B → G : A ⟶ B
39 38 ffvelcdmda ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → G ⁡ x ∈ B
40 fof ⊢ F : A ⟶ onto B → F : A ⟶ B
41 40 adantr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B → F : A ⟶ B
42 41 ffvelcdmda ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → F ⁡ x ∈ B
43 resieq ⊢ G ⁡ x ∈ B ∧ F ⁡ x ∈ B → G ⁡ x I ↾ B F ⁡ x ↔ G ⁡ x = F ⁡ x
44 39 42 43 syl2anc ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ x ∈ A → G ⁡ x I ↾ B F ⁡ x ↔ G ⁡ x = F ⁡ x
45 44 adantlr ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B ∧ x ∈ A → G ⁡ x I ↾ B F ⁡ x ↔ G ⁡ x = F ⁡ x
46 36 45 mpbid ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B ∧ x ∈ A → G ⁡ x = F ⁡ x
47 46 eqcomd ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B ∧ x ∈ A → F ⁡ x = G ⁡ x
48 8 10 47 eqfnfvd ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B ∧ F ∘ G -1 = I ↾ B → F = G
49 48 ex ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B → F ∘ G -1 = I ↾ B → F = G
50 6 49 impbid ⊢ F : A ⟶ onto B ∧ G : A ⟶ onto B → F = G ↔ F ∘ G -1 = I ↾ B