Step | Hyp | Ref
| Expression |
1 | | ofco.3 |
. . . 4
⊢ (𝜑 → 𝐻:𝐷⟶𝐶) |
2 | 1 | ffvelrnda 6958 |
. . 3
⊢ ((𝜑 ∧ 𝑥 ∈ 𝐷) → (𝐻‘𝑥) ∈ 𝐶) |
3 | 1 | feqmptd 6834 |
. . 3
⊢ (𝜑 → 𝐻 = (𝑥 ∈ 𝐷 ↦ (𝐻‘𝑥))) |
4 | | ofco.1 |
. . . 4
⊢ (𝜑 → 𝐹 Fn 𝐴) |
5 | | ofco.2 |
. . . 4
⊢ (𝜑 → 𝐺 Fn 𝐵) |
6 | | ofco.4 |
. . . 4
⊢ (𝜑 → 𝐴 ∈ 𝑉) |
7 | | ofco.5 |
. . . 4
⊢ (𝜑 → 𝐵 ∈ 𝑊) |
8 | | ofco.7 |
. . . 4
⊢ (𝐴 ∩ 𝐵) = 𝐶 |
9 | | eqidd 2741 |
. . . 4
⊢ ((𝜑 ∧ 𝑦 ∈ 𝐴) → (𝐹‘𝑦) = (𝐹‘𝑦)) |
10 | | eqidd 2741 |
. . . 4
⊢ ((𝜑 ∧ 𝑦 ∈ 𝐵) → (𝐺‘𝑦) = (𝐺‘𝑦)) |
11 | 4, 5, 6, 7, 8, 9, 10 | offval 7536 |
. . 3
⊢ (𝜑 → (𝐹 ∘f 𝑅𝐺) = (𝑦 ∈ 𝐶 ↦ ((𝐹‘𝑦)𝑅(𝐺‘𝑦)))) |
12 | | fveq2 6771 |
. . . 4
⊢ (𝑦 = (𝐻‘𝑥) → (𝐹‘𝑦) = (𝐹‘(𝐻‘𝑥))) |
13 | | fveq2 6771 |
. . . 4
⊢ (𝑦 = (𝐻‘𝑥) → (𝐺‘𝑦) = (𝐺‘(𝐻‘𝑥))) |
14 | 12, 13 | oveq12d 7289 |
. . 3
⊢ (𝑦 = (𝐻‘𝑥) → ((𝐹‘𝑦)𝑅(𝐺‘𝑦)) = ((𝐹‘(𝐻‘𝑥))𝑅(𝐺‘(𝐻‘𝑥)))) |
15 | 2, 3, 11, 14 | fmptco 6998 |
. 2
⊢ (𝜑 → ((𝐹 ∘f 𝑅𝐺) ∘ 𝐻) = (𝑥 ∈ 𝐷 ↦ ((𝐹‘(𝐻‘𝑥))𝑅(𝐺‘(𝐻‘𝑥))))) |
16 | | inss1 4168 |
. . . . . 6
⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴 |
17 | 8, 16 | eqsstrri 3961 |
. . . . 5
⊢ 𝐶 ⊆ 𝐴 |
18 | | fss 6615 |
. . . . 5
⊢ ((𝐻:𝐷⟶𝐶 ∧ 𝐶 ⊆ 𝐴) → 𝐻:𝐷⟶𝐴) |
19 | 1, 17, 18 | sylancl 586 |
. . . 4
⊢ (𝜑 → 𝐻:𝐷⟶𝐴) |
20 | | fnfco 6637 |
. . . 4
⊢ ((𝐹 Fn 𝐴 ∧ 𝐻:𝐷⟶𝐴) → (𝐹 ∘ 𝐻) Fn 𝐷) |
21 | 4, 19, 20 | syl2anc 584 |
. . 3
⊢ (𝜑 → (𝐹 ∘ 𝐻) Fn 𝐷) |
22 | | inss2 4169 |
. . . . . 6
⊢ (𝐴 ∩ 𝐵) ⊆ 𝐵 |
23 | 8, 22 | eqsstrri 3961 |
. . . . 5
⊢ 𝐶 ⊆ 𝐵 |
24 | | fss 6615 |
. . . . 5
⊢ ((𝐻:𝐷⟶𝐶 ∧ 𝐶 ⊆ 𝐵) → 𝐻:𝐷⟶𝐵) |
25 | 1, 23, 24 | sylancl 586 |
. . . 4
⊢ (𝜑 → 𝐻:𝐷⟶𝐵) |
26 | | fnfco 6637 |
. . . 4
⊢ ((𝐺 Fn 𝐵 ∧ 𝐻:𝐷⟶𝐵) → (𝐺 ∘ 𝐻) Fn 𝐷) |
27 | 5, 25, 26 | syl2anc 584 |
. . 3
⊢ (𝜑 → (𝐺 ∘ 𝐻) Fn 𝐷) |
28 | | ofco.6 |
. . 3
⊢ (𝜑 → 𝐷 ∈ 𝑋) |
29 | | inidm 4158 |
. . 3
⊢ (𝐷 ∩ 𝐷) = 𝐷 |
30 | 1 | ffnd 6599 |
. . . 4
⊢ (𝜑 → 𝐻 Fn 𝐷) |
31 | | fvco2 6862 |
. . . 4
⊢ ((𝐻 Fn 𝐷 ∧ 𝑥 ∈ 𝐷) → ((𝐹 ∘ 𝐻)‘𝑥) = (𝐹‘(𝐻‘𝑥))) |
32 | 30, 31 | sylan 580 |
. . 3
⊢ ((𝜑 ∧ 𝑥 ∈ 𝐷) → ((𝐹 ∘ 𝐻)‘𝑥) = (𝐹‘(𝐻‘𝑥))) |
33 | | fvco2 6862 |
. . . 4
⊢ ((𝐻 Fn 𝐷 ∧ 𝑥 ∈ 𝐷) → ((𝐺 ∘ 𝐻)‘𝑥) = (𝐺‘(𝐻‘𝑥))) |
34 | 30, 33 | sylan 580 |
. . 3
⊢ ((𝜑 ∧ 𝑥 ∈ 𝐷) → ((𝐺 ∘ 𝐻)‘𝑥) = (𝐺‘(𝐻‘𝑥))) |
35 | 21, 27, 28, 28, 29, 32, 34 | offval 7536 |
. 2
⊢ (𝜑 → ((𝐹 ∘ 𝐻) ∘f 𝑅(𝐺 ∘ 𝐻)) = (𝑥 ∈ 𝐷 ↦ ((𝐹‘(𝐻‘𝑥))𝑅(𝐺‘(𝐻‘𝑥))))) |
36 | 15, 35 | eqtr4d 2783 |
1
⊢ (𝜑 → ((𝐹 ∘f 𝑅𝐺) ∘ 𝐻) = ((𝐹 ∘ 𝐻) ∘f 𝑅(𝐺 ∘ 𝐻))) |