Proof of Theorem f1un
Step | Hyp | Ref
| Expression |
1 | | f1f 6670 |
. . . 4
⊢ (𝐹:𝐴–1-1→𝐵 → 𝐹:𝐴⟶𝐵) |
2 | 1 | frnd 6608 |
. . 3
⊢ (𝐹:𝐴–1-1→𝐵 → ran 𝐹 ⊆ 𝐵) |
3 | | f1f 6670 |
. . . 4
⊢ (𝐺:𝐶–1-1→𝐷 → 𝐺:𝐶⟶𝐷) |
4 | 3 | frnd 6608 |
. . 3
⊢ (𝐺:𝐶–1-1→𝐷 → ran 𝐺 ⊆ 𝐷) |
5 | | unss12 4116 |
. . 3
⊢ ((ran
𝐹 ⊆ 𝐵 ∧ ran 𝐺 ⊆ 𝐷) → (ran 𝐹 ∪ ran 𝐺) ⊆ (𝐵 ∪ 𝐷)) |
6 | 2, 4, 5 | syl2an 596 |
. 2
⊢ ((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) → (ran 𝐹 ∪ ran 𝐺) ⊆ (𝐵 ∪ 𝐷)) |
7 | | f1f1orn 6727 |
. . . . 5
⊢ (𝐹:𝐴–1-1→𝐵 → 𝐹:𝐴–1-1-onto→ran
𝐹) |
8 | | f1f1orn 6727 |
. . . . 5
⊢ (𝐺:𝐶–1-1→𝐷 → 𝐺:𝐶–1-1-onto→ran
𝐺) |
9 | 7, 8 | anim12i 613 |
. . . 4
⊢ ((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) → (𝐹:𝐴–1-1-onto→ran
𝐹 ∧ 𝐺:𝐶–1-1-onto→ran
𝐺)) |
10 | | simprl 768 |
. . . . 5
⊢ (((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐴 ∩ 𝐶) = ∅) |
11 | | ss2in 4170 |
. . . . . . . 8
⊢ ((ran
𝐹 ⊆ 𝐵 ∧ ran 𝐺 ⊆ 𝐷) → (ran 𝐹 ∩ ran 𝐺) ⊆ (𝐵 ∩ 𝐷)) |
12 | 2, 4, 11 | syl2an 596 |
. . . . . . 7
⊢ ((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) → (ran 𝐹 ∩ ran 𝐺) ⊆ (𝐵 ∩ 𝐷)) |
13 | | sseq0 4333 |
. . . . . . 7
⊢ (((ran
𝐹 ∩ ran 𝐺) ⊆ (𝐵 ∩ 𝐷) ∧ (𝐵 ∩ 𝐷) = ∅) → (ran 𝐹 ∩ ran 𝐺) = ∅) |
14 | 12, 13 | sylan 580 |
. . . . . 6
⊢ (((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) ∧ (𝐵 ∩ 𝐷) = ∅) → (ran 𝐹 ∩ ran 𝐺) = ∅) |
15 | 14 | adantrl 713 |
. . . . 5
⊢ (((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (ran 𝐹 ∩ ran 𝐺) = ∅) |
16 | 10, 15 | jca 512 |
. . . 4
⊢ (((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → ((𝐴 ∩ 𝐶) = ∅ ∧ (ran 𝐹 ∩ ran 𝐺) = ∅)) |
17 | | f1oun 6735 |
. . . 4
⊢ (((𝐹:𝐴–1-1-onto→ran
𝐹 ∧ 𝐺:𝐶–1-1-onto→ran
𝐺) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (ran 𝐹 ∩ ran 𝐺) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1-onto→(ran
𝐹 ∪ ran 𝐺)) |
18 | 9, 16, 17 | syl2an2r 682 |
. . 3
⊢ (((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1-onto→(ran
𝐹 ∪ ran 𝐺)) |
19 | | f1of1 6715 |
. . 3
⊢ ((𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1-onto→(ran
𝐹 ∪ ran 𝐺) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1→(ran 𝐹 ∪ ran 𝐺)) |
20 | 18, 19 | syl 17 |
. 2
⊢ (((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1→(ran 𝐹 ∪ ran 𝐺)) |
21 | | f1ss 6676 |
. . 3
⊢ (((𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1→(ran 𝐹 ∪ ran 𝐺) ∧ (ran 𝐹 ∪ ran 𝐺) ⊆ (𝐵 ∪ 𝐷)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1→(𝐵 ∪ 𝐷)) |
22 | 21 | ancoms 459 |
. 2
⊢ (((ran
𝐹 ∪ ran 𝐺) ⊆ (𝐵 ∪ 𝐷) ∧ (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1→(ran 𝐹 ∪ ran 𝐺)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1→(𝐵 ∪ 𝐷)) |
23 | 6, 20, 22 | syl2an2r 682 |
1
⊢ (((𝐹:𝐴–1-1→𝐵 ∧ 𝐺:𝐶–1-1→𝐷) ∧ ((𝐴 ∩ 𝐶) = ∅ ∧ (𝐵 ∩ 𝐷) = ∅)) → (𝐹 ∪ 𝐺):(𝐴 ∪ 𝐶)–1-1→(𝐵 ∪ 𝐷)) |