Proof of Theorem preiman0
Step | Hyp | Ref
| Expression |
1 | | df-rn 5599 |
. . . . . 6
⊢ ran 𝐹 = dom ◡𝐹 |
2 | 1 | ineq1i 4147 |
. . . . 5
⊢ (ran
𝐹 ∩ (𝐴 ∩ ran 𝐹)) = (dom ◡𝐹 ∩ (𝐴 ∩ ran 𝐹)) |
3 | | df-ss 3908 |
. . . . . . . . 9
⊢ (𝐴 ⊆ ran 𝐹 ↔ (𝐴 ∩ ran 𝐹) = 𝐴) |
4 | 3 | biimpi 215 |
. . . . . . . 8
⊢ (𝐴 ⊆ ran 𝐹 → (𝐴 ∩ ran 𝐹) = 𝐴) |
5 | 4 | ineq2d 4151 |
. . . . . . 7
⊢ (𝐴 ⊆ ran 𝐹 → (ran 𝐹 ∩ (𝐴 ∩ ran 𝐹)) = (ran 𝐹 ∩ 𝐴)) |
6 | | sseqin2 4154 |
. . . . . . . 8
⊢ (𝐴 ⊆ ran 𝐹 ↔ (ran 𝐹 ∩ 𝐴) = 𝐴) |
7 | 6 | biimpi 215 |
. . . . . . 7
⊢ (𝐴 ⊆ ran 𝐹 → (ran 𝐹 ∩ 𝐴) = 𝐴) |
8 | 5, 7 | eqtrd 2779 |
. . . . . 6
⊢ (𝐴 ⊆ ran 𝐹 → (ran 𝐹 ∩ (𝐴 ∩ ran 𝐹)) = 𝐴) |
9 | 8 | 3ad2ant2 1132 |
. . . . 5
⊢ ((Fun
𝐹 ∧ 𝐴 ⊆ ran 𝐹 ∧ (◡𝐹 “ 𝐴) = ∅) → (ran 𝐹 ∩ (𝐴 ∩ ran 𝐹)) = 𝐴) |
10 | | fimacnvinrn 6943 |
. . . . . . . . 9
⊢ (Fun
𝐹 → (◡𝐹 “ 𝐴) = (◡𝐹 “ (𝐴 ∩ ran 𝐹))) |
11 | 10 | eqeq1d 2741 |
. . . . . . . 8
⊢ (Fun
𝐹 → ((◡𝐹 “ 𝐴) = ∅ ↔ (◡𝐹 “ (𝐴 ∩ ran 𝐹)) = ∅)) |
12 | 11 | biimpa 476 |
. . . . . . 7
⊢ ((Fun
𝐹 ∧ (◡𝐹 “ 𝐴) = ∅) → (◡𝐹 “ (𝐴 ∩ ran 𝐹)) = ∅) |
13 | 12 | 3adant2 1129 |
. . . . . 6
⊢ ((Fun
𝐹 ∧ 𝐴 ⊆ ran 𝐹 ∧ (◡𝐹 “ 𝐴) = ∅) → (◡𝐹 “ (𝐴 ∩ ran 𝐹)) = ∅) |
14 | | imadisj 5985 |
. . . . . 6
⊢ ((◡𝐹 “ (𝐴 ∩ ran 𝐹)) = ∅ ↔ (dom ◡𝐹 ∩ (𝐴 ∩ ran 𝐹)) = ∅) |
15 | 13, 14 | sylib 217 |
. . . . 5
⊢ ((Fun
𝐹 ∧ 𝐴 ⊆ ran 𝐹 ∧ (◡𝐹 “ 𝐴) = ∅) → (dom ◡𝐹 ∩ (𝐴 ∩ ran 𝐹)) = ∅) |
16 | 2, 9, 15 | 3eqtr3a 2803 |
. . . 4
⊢ ((Fun
𝐹 ∧ 𝐴 ⊆ ran 𝐹 ∧ (◡𝐹 “ 𝐴) = ∅) → 𝐴 = ∅) |
17 | 16 | 3expia 1119 |
. . 3
⊢ ((Fun
𝐹 ∧ 𝐴 ⊆ ran 𝐹) → ((◡𝐹 “ 𝐴) = ∅ → 𝐴 = ∅)) |
18 | 17 | necon3d 2965 |
. 2
⊢ ((Fun
𝐹 ∧ 𝐴 ⊆ ran 𝐹) → (𝐴 ≠ ∅ → (◡𝐹 “ 𝐴) ≠ ∅)) |
19 | 18 | 3impia 1115 |
1
⊢ ((Fun
𝐹 ∧ 𝐴 ⊆ ran 𝐹 ∧ 𝐴 ≠ ∅) → (◡𝐹 “ 𝐴) ≠ ∅) |