Step | Hyp | Ref
| Expression |
1 | | off.2 |
. . . . 5
⊢ (𝜑 → 𝐹:𝐴⟶𝑆) |
2 | | off.6 |
. . . . . . 7
⊢ (𝐴 ∩ 𝐵) = 𝐶 |
3 | | inss1 3221 |
. . . . . . 7
⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴 |
4 | 2, 3 | eqsstr3i 3058 |
. . . . . 6
⊢ 𝐶 ⊆ 𝐴 |
5 | 4 | sseli 3022 |
. . . . 5
⊢ (𝑧 ∈ 𝐶 → 𝑧 ∈ 𝐴) |
6 | | ffvelrn 5446 |
. . . . 5
⊢ ((𝐹:𝐴⟶𝑆 ∧ 𝑧 ∈ 𝐴) → (𝐹‘𝑧) ∈ 𝑆) |
7 | 1, 5, 6 | syl2an 284 |
. . . 4
⊢ ((𝜑 ∧ 𝑧 ∈ 𝐶) → (𝐹‘𝑧) ∈ 𝑆) |
8 | | off.3 |
. . . . 5
⊢ (𝜑 → 𝐺:𝐵⟶𝑇) |
9 | | inss2 3222 |
. . . . . . 7
⊢ (𝐴 ∩ 𝐵) ⊆ 𝐵 |
10 | 2, 9 | eqsstr3i 3058 |
. . . . . 6
⊢ 𝐶 ⊆ 𝐵 |
11 | 10 | sseli 3022 |
. . . . 5
⊢ (𝑧 ∈ 𝐶 → 𝑧 ∈ 𝐵) |
12 | | ffvelrn 5446 |
. . . . 5
⊢ ((𝐺:𝐵⟶𝑇 ∧ 𝑧 ∈ 𝐵) → (𝐺‘𝑧) ∈ 𝑇) |
13 | 8, 11, 12 | syl2an 284 |
. . . 4
⊢ ((𝜑 ∧ 𝑧 ∈ 𝐶) → (𝐺‘𝑧) ∈ 𝑇) |
14 | | off.1 |
. . . . . 6
⊢ ((𝜑 ∧ (𝑥 ∈ 𝑆 ∧ 𝑦 ∈ 𝑇)) → (𝑥𝑅𝑦) ∈ 𝑈) |
15 | 14 | ralrimivva 2456 |
. . . . 5
⊢ (𝜑 → ∀𝑥 ∈ 𝑆 ∀𝑦 ∈ 𝑇 (𝑥𝑅𝑦) ∈ 𝑈) |
16 | 15 | adantr 271 |
. . . 4
⊢ ((𝜑 ∧ 𝑧 ∈ 𝐶) → ∀𝑥 ∈ 𝑆 ∀𝑦 ∈ 𝑇 (𝑥𝑅𝑦) ∈ 𝑈) |
17 | | oveq1 5673 |
. . . . . 6
⊢ (𝑥 = (𝐹‘𝑧) → (𝑥𝑅𝑦) = ((𝐹‘𝑧)𝑅𝑦)) |
18 | 17 | eleq1d 2157 |
. . . . 5
⊢ (𝑥 = (𝐹‘𝑧) → ((𝑥𝑅𝑦) ∈ 𝑈 ↔ ((𝐹‘𝑧)𝑅𝑦) ∈ 𝑈)) |
19 | | oveq2 5674 |
. . . . . 6
⊢ (𝑦 = (𝐺‘𝑧) → ((𝐹‘𝑧)𝑅𝑦) = ((𝐹‘𝑧)𝑅(𝐺‘𝑧))) |
20 | 19 | eleq1d 2157 |
. . . . 5
⊢ (𝑦 = (𝐺‘𝑧) → (((𝐹‘𝑧)𝑅𝑦) ∈ 𝑈 ↔ ((𝐹‘𝑧)𝑅(𝐺‘𝑧)) ∈ 𝑈)) |
21 | 18, 20 | rspc2va 2736 |
. . . 4
⊢ ((((𝐹‘𝑧) ∈ 𝑆 ∧ (𝐺‘𝑧) ∈ 𝑇) ∧ ∀𝑥 ∈ 𝑆 ∀𝑦 ∈ 𝑇 (𝑥𝑅𝑦) ∈ 𝑈) → ((𝐹‘𝑧)𝑅(𝐺‘𝑧)) ∈ 𝑈) |
22 | 7, 13, 16, 21 | syl21anc 1174 |
. . 3
⊢ ((𝜑 ∧ 𝑧 ∈ 𝐶) → ((𝐹‘𝑧)𝑅(𝐺‘𝑧)) ∈ 𝑈) |
23 | | eqid 2089 |
. . 3
⊢ (𝑧 ∈ 𝐶 ↦ ((𝐹‘𝑧)𝑅(𝐺‘𝑧))) = (𝑧 ∈ 𝐶 ↦ ((𝐹‘𝑧)𝑅(𝐺‘𝑧))) |
24 | 22, 23 | fmptd 5466 |
. 2
⊢ (𝜑 → (𝑧 ∈ 𝐶 ↦ ((𝐹‘𝑧)𝑅(𝐺‘𝑧))):𝐶⟶𝑈) |
25 | | ffn 5174 |
. . . . 5
⊢ (𝐹:𝐴⟶𝑆 → 𝐹 Fn 𝐴) |
26 | 1, 25 | syl 14 |
. . . 4
⊢ (𝜑 → 𝐹 Fn 𝐴) |
27 | | ffn 5174 |
. . . . 5
⊢ (𝐺:𝐵⟶𝑇 → 𝐺 Fn 𝐵) |
28 | 8, 27 | syl 14 |
. . . 4
⊢ (𝜑 → 𝐺 Fn 𝐵) |
29 | | off.4 |
. . . 4
⊢ (𝜑 → 𝐴 ∈ 𝑉) |
30 | | off.5 |
. . . 4
⊢ (𝜑 → 𝐵 ∈ 𝑊) |
31 | | eqidd 2090 |
. . . 4
⊢ ((𝜑 ∧ 𝑧 ∈ 𝐴) → (𝐹‘𝑧) = (𝐹‘𝑧)) |
32 | | eqidd 2090 |
. . . 4
⊢ ((𝜑 ∧ 𝑧 ∈ 𝐵) → (𝐺‘𝑧) = (𝐺‘𝑧)) |
33 | 26, 28, 29, 30, 2, 31, 32 | offval 5877 |
. . 3
⊢ (𝜑 → (𝐹 ∘𝑓 𝑅𝐺) = (𝑧 ∈ 𝐶 ↦ ((𝐹‘𝑧)𝑅(𝐺‘𝑧)))) |
34 | 33 | feq1d 5162 |
. 2
⊢ (𝜑 → ((𝐹 ∘𝑓 𝑅𝐺):𝐶⟶𝑈 ↔ (𝑧 ∈ 𝐶 ↦ ((𝐹‘𝑧)𝑅(𝐺‘𝑧))):𝐶⟶𝑈)) |
35 | 24, 34 | mpbird 166 |
1
⊢ (𝜑 → (𝐹 ∘𝑓 𝑅𝐺):𝐶⟶𝑈) |