Metamath Proof Explorer


Theorem fvcofneq

Description: The values of two function compositions are equal if the values of the composed functions are pairwise equal. (Contributed by AV, 26-Jan-2019)

Ref Expression
Assertion fvcofneq ⊢ G Fn A ∧ K Fn B → X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → F ∘ G ⁡ X = H ∘ K ⁡ X

Proof

Step Hyp Ref Expression
1 simpl ⊢ G Fn A ∧ K Fn B → G Fn A
2 elinel1 ⊢ X ∈ A ∩ B → X ∈ A
3 2 3ad2ant1 ⊢ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → X ∈ A
4 fvco2 ⊢ G Fn A ∧ X ∈ A → F ∘ G ⁡ X = F ⁡ G ⁡ X
5 1 3 4 syl2an ⊢ G Fn A ∧ K Fn B ∧ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → F ∘ G ⁡ X = F ⁡ G ⁡ X
6 simpr ⊢ G Fn A ∧ K Fn B → K Fn B
7 elinel2 ⊢ X ∈ A ∩ B → X ∈ B
8 7 3ad2ant1 ⊢ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → X ∈ B
9 fvco2 ⊢ K Fn B ∧ X ∈ B → H ∘ K ⁡ X = H ⁡ K ⁡ X
10 6 8 9 syl2an ⊢ G Fn A ∧ K Fn B ∧ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → H ∘ K ⁡ X = H ⁡ K ⁡ X
11 fveq2 ⊢ K ⁡ X = G ⁡ X → H ⁡ K ⁡ X = H ⁡ G ⁡ X
12 11 eqcoms ⊢ G ⁡ X = K ⁡ X → H ⁡ K ⁡ X = H ⁡ G ⁡ X
13 12 3ad2ant2 ⊢ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → H ⁡ K ⁡ X = H ⁡ G ⁡ X
14 13 adantl ⊢ G Fn A ∧ K Fn B ∧ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → H ⁡ K ⁡ X = H ⁡ G ⁡ X
15 id ⊢ G Fn A → G Fn A
16 fnfvelrn ⊢ G Fn A ∧ X ∈ A → G ⁡ X ∈ ran ⁡ G
17 15 2 16 syl2anr ⊢ X ∈ A ∩ B ∧ G Fn A → G ⁡ X ∈ ran ⁡ G
18 17 ex ⊢ X ∈ A ∩ B → G Fn A → G ⁡ X ∈ ran ⁡ G
19 id ⊢ K Fn B → K Fn B
20 fnfvelrn ⊢ K Fn B ∧ X ∈ B → K ⁡ X ∈ ran ⁡ K
21 19 7 20 syl2anr ⊢ X ∈ A ∩ B ∧ K Fn B → K ⁡ X ∈ ran ⁡ K
22 21 ex ⊢ X ∈ A ∩ B → K Fn B → K ⁡ X ∈ ran ⁡ K
23 18 22 anim12d ⊢ X ∈ A ∩ B → G Fn A ∧ K Fn B → G ⁡ X ∈ ran ⁡ G ∧ K ⁡ X ∈ ran ⁡ K
24 eleq1 ⊢ K ⁡ X = G ⁡ X → K ⁡ X ∈ ran ⁡ K ↔ G ⁡ X ∈ ran ⁡ K
25 24 eqcoms ⊢ G ⁡ X = K ⁡ X → K ⁡ X ∈ ran ⁡ K ↔ G ⁡ X ∈ ran ⁡ K
26 25 anbi2d ⊢ G ⁡ X = K ⁡ X → G ⁡ X ∈ ran ⁡ G ∧ K ⁡ X ∈ ran ⁡ K ↔ G ⁡ X ∈ ran ⁡ G ∧ G ⁡ X ∈ ran ⁡ K
27 elin ⊢ G ⁡ X ∈ ran ⁡ G ∩ ran ⁡ K ↔ G ⁡ X ∈ ran ⁡ G ∧ G ⁡ X ∈ ran ⁡ K
28 27 biimpri ⊢ G ⁡ X ∈ ran ⁡ G ∧ G ⁡ X ∈ ran ⁡ K → G ⁡ X ∈ ran ⁡ G ∩ ran ⁡ K
29 26 28 biimtrdi ⊢ G ⁡ X = K ⁡ X → G ⁡ X ∈ ran ⁡ G ∧ K ⁡ X ∈ ran ⁡ K → G ⁡ X ∈ ran ⁡ G ∩ ran ⁡ K
30 23 29 sylan9 ⊢ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X → G Fn A ∧ K Fn B → G ⁡ X ∈ ran ⁡ G ∩ ran ⁡ K
31 fveq2 ⊢ x = G ⁡ X → F ⁡ x = F ⁡ G ⁡ X
32 fveq2 ⊢ x = G ⁡ X → H ⁡ x = H ⁡ G ⁡ X
33 31 32 eqeq12d ⊢ x = G ⁡ X → F ⁡ x = H ⁡ x ↔ F ⁡ G ⁡ X = H ⁡ G ⁡ X
34 33 rspcva ⊢ G ⁡ X ∈ ran ⁡ G ∩ ran ⁡ K ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → F ⁡ G ⁡ X = H ⁡ G ⁡ X
35 34 eqcomd ⊢ G ⁡ X ∈ ran ⁡ G ∩ ran ⁡ K ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → H ⁡ G ⁡ X = F ⁡ G ⁡ X
36 35 ex ⊢ G ⁡ X ∈ ran ⁡ G ∩ ran ⁡ K → ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → H ⁡ G ⁡ X = F ⁡ G ⁡ X
37 30 36 syl6 ⊢ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X → G Fn A ∧ K Fn B → ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → H ⁡ G ⁡ X = F ⁡ G ⁡ X
38 37 com23 ⊢ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X → ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → G Fn A ∧ K Fn B → H ⁡ G ⁡ X = F ⁡ G ⁡ X
39 38 3impia ⊢ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → G Fn A ∧ K Fn B → H ⁡ G ⁡ X = F ⁡ G ⁡ X
40 39 impcom ⊢ G Fn A ∧ K Fn B ∧ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → H ⁡ G ⁡ X = F ⁡ G ⁡ X
41 10 14 40 3eqtrrd ⊢ G Fn A ∧ K Fn B ∧ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → F ⁡ G ⁡ X = H ∘ K ⁡ X
42 5 41 eqtrd ⊢ G Fn A ∧ K Fn B ∧ X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → F ∘ G ⁡ X = H ∘ K ⁡ X
43 42 ex ⊢ G Fn A ∧ K Fn B → X ∈ A ∩ B ∧ G ⁡ X = K ⁡ X ∧ ∀ x ∈ ran ⁡ G ∩ ran ⁡ K F ⁡ x = H ⁡ x → F ∘ G ⁡ X = H ∘ K ⁡ X