| Step | Hyp | Ref
| Expression |
| 1 | | nlfnval 28740 |
. . . . . 6
⊢ (𝑇: ℋ⟶ℂ →
(null‘𝑇) = (◡𝑇 “ {0})) |
| 2 | | cnvimass 5485 |
. . . . . 6
⊢ (◡𝑇 “ {0}) ⊆ dom 𝑇 |
| 3 | 1, 2 | syl6eqss 3655 |
. . . . 5
⊢ (𝑇: ℋ⟶ℂ →
(null‘𝑇) ⊆ dom
𝑇) |
| 4 | | fdm 6051 |
. . . . 5
⊢ (𝑇: ℋ⟶ℂ →
dom 𝑇 =
ℋ) |
| 5 | 3, 4 | sseqtrd 3641 |
. . . 4
⊢ (𝑇: ℋ⟶ℂ →
(null‘𝑇) ⊆
ℋ) |
| 6 | 5 | sseld 3602 |
. . 3
⊢ (𝑇: ℋ⟶ℂ →
(𝐴 ∈ (null‘𝑇) → 𝐴 ∈ ℋ)) |
| 7 | 6 | pm4.71rd 667 |
. 2
⊢ (𝑇: ℋ⟶ℂ →
(𝐴 ∈ (null‘𝑇) ↔ (𝐴 ∈ ℋ ∧ 𝐴 ∈ (null‘𝑇)))) |
| 8 | 1 | eleq2d 2687 |
. . . . 5
⊢ (𝑇: ℋ⟶ℂ →
(𝐴 ∈ (null‘𝑇) ↔ 𝐴 ∈ (◡𝑇 “ {0}))) |
| 9 | 8 | adantr 481 |
. . . 4
⊢ ((𝑇: ℋ⟶ℂ ∧
𝐴 ∈ ℋ) →
(𝐴 ∈ (null‘𝑇) ↔ 𝐴 ∈ (◡𝑇 “ {0}))) |
| 10 | | ffn 6045 |
. . . . 5
⊢ (𝑇: ℋ⟶ℂ →
𝑇 Fn
ℋ) |
| 11 | | eleq1 2689 |
. . . . . . . 8
⊢ (𝑥 = 𝐴 → (𝑥 ∈ (◡𝑇 “ {0}) ↔ 𝐴 ∈ (◡𝑇 “ {0}))) |
| 12 | | fveq2 6191 |
. . . . . . . . 9
⊢ (𝑥 = 𝐴 → (𝑇‘𝑥) = (𝑇‘𝐴)) |
| 13 | 12 | eqeq1d 2624 |
. . . . . . . 8
⊢ (𝑥 = 𝐴 → ((𝑇‘𝑥) = 0 ↔ (𝑇‘𝐴) = 0)) |
| 14 | 11, 13 | bibi12d 335 |
. . . . . . 7
⊢ (𝑥 = 𝐴 → ((𝑥 ∈ (◡𝑇 “ {0}) ↔ (𝑇‘𝑥) = 0) ↔ (𝐴 ∈ (◡𝑇 “ {0}) ↔ (𝑇‘𝐴) = 0))) |
| 15 | 14 | imbi2d 330 |
. . . . . 6
⊢ (𝑥 = 𝐴 → ((𝑇 Fn ℋ → (𝑥 ∈ (◡𝑇 “ {0}) ↔ (𝑇‘𝑥) = 0)) ↔ (𝑇 Fn ℋ → (𝐴 ∈ (◡𝑇 “ {0}) ↔ (𝑇‘𝐴) = 0)))) |
| 16 | | fnbrfvb 6236 |
. . . . . . . 8
⊢ ((𝑇 Fn ℋ ∧ 𝑥 ∈ ℋ) → ((𝑇‘𝑥) = 0 ↔ 𝑥𝑇0)) |
| 17 | | 0cn 10032 |
. . . . . . . . 9
⊢ 0 ∈
ℂ |
| 18 | | vex 3203 |
. . . . . . . . . 10
⊢ 𝑥 ∈ V |
| 19 | 18 | eliniseg 5494 |
. . . . . . . . 9
⊢ (0 ∈
ℂ → (𝑥 ∈
(◡𝑇 “ {0}) ↔ 𝑥𝑇0)) |
| 20 | 17, 19 | ax-mp 5 |
. . . . . . . 8
⊢ (𝑥 ∈ (◡𝑇 “ {0}) ↔ 𝑥𝑇0) |
| 21 | 16, 20 | syl6rbbr 279 |
. . . . . . 7
⊢ ((𝑇 Fn ℋ ∧ 𝑥 ∈ ℋ) → (𝑥 ∈ (◡𝑇 “ {0}) ↔ (𝑇‘𝑥) = 0)) |
| 22 | 21 | expcom 451 |
. . . . . 6
⊢ (𝑥 ∈ ℋ → (𝑇 Fn ℋ → (𝑥 ∈ (◡𝑇 “ {0}) ↔ (𝑇‘𝑥) = 0))) |
| 23 | 15, 22 | vtoclga 3272 |
. . . . 5
⊢ (𝐴 ∈ ℋ → (𝑇 Fn ℋ → (𝐴 ∈ (◡𝑇 “ {0}) ↔ (𝑇‘𝐴) = 0))) |
| 24 | 10, 23 | mpan9 486 |
. . . 4
⊢ ((𝑇: ℋ⟶ℂ ∧
𝐴 ∈ ℋ) →
(𝐴 ∈ (◡𝑇 “ {0}) ↔ (𝑇‘𝐴) = 0)) |
| 25 | 9, 24 | bitrd 268 |
. . 3
⊢ ((𝑇: ℋ⟶ℂ ∧
𝐴 ∈ ℋ) →
(𝐴 ∈ (null‘𝑇) ↔ (𝑇‘𝐴) = 0)) |
| 26 | 25 | pm5.32da 673 |
. 2
⊢ (𝑇: ℋ⟶ℂ →
((𝐴 ∈ ℋ ∧
𝐴 ∈ (null‘𝑇)) ↔ (𝐴 ∈ ℋ ∧ (𝑇‘𝐴) = 0))) |
| 27 | 7, 26 | bitrd 268 |
1
⊢ (𝑇: ℋ⟶ℂ →
(𝐴 ∈ (null‘𝑇) ↔ (𝐴 ∈ ℋ ∧ (𝑇‘𝐴) = 0))) |