Proof of Theorem ceqsralt
Step | Hyp | Ref
| Expression |
1 | | df-ral 2620 |
. . . 4
⊢ (∀x ∈ B (x = A →
φ) ↔ ∀x(x ∈ B → (x =
A → φ))) |
2 | | eleq1 2413 |
. . . . . . . . 9
⊢ (x = A →
(x ∈
B ↔ A ∈ B)) |
3 | 2 | pm5.32ri 619 |
. . . . . . . 8
⊢ ((x ∈ B ∧ x = A) ↔
(A ∈
B ∧
x = A)) |
4 | 3 | imbi1i 315 |
. . . . . . 7
⊢ (((x ∈ B ∧ x = A) →
φ) ↔ ((A ∈ B ∧ x = A) →
φ)) |
5 | | impexp 433 |
. . . . . . 7
⊢ (((x ∈ B ∧ x = A) →
φ) ↔ (x ∈ B → (x =
A → φ))) |
6 | | impexp 433 |
. . . . . . 7
⊢ (((A ∈ B ∧ x = A) →
φ) ↔ (A ∈ B → (x =
A → φ))) |
7 | 4, 5, 6 | 3bitr3i 266 |
. . . . . 6
⊢ ((x ∈ B → (x =
A → φ)) ↔ (A ∈ B → (x =
A → φ))) |
8 | 7 | albii 1566 |
. . . . 5
⊢ (∀x(x ∈ B → (x =
A → φ)) ↔ ∀x(A ∈ B → (x =
A → φ))) |
9 | 8 | a1i 10 |
. . . 4
⊢ ((Ⅎxψ ∧ ∀x(x = A → (φ
↔ ψ)) ∧ A ∈ B) →
(∀x(x ∈ B →
(x = A
→ φ)) ↔ ∀x(A ∈ B → (x =
A → φ)))) |
10 | 1, 9 | syl5bb 248 |
. . 3
⊢ ((Ⅎxψ ∧ ∀x(x = A → (φ
↔ ψ)) ∧ A ∈ B) →
(∀x
∈ B
(x = A
→ φ) ↔ ∀x(A ∈ B → (x =
A → φ)))) |
11 | | 19.21v 1890 |
. . 3
⊢ (∀x(A ∈ B → (x =
A → φ)) ↔ (A ∈ B → ∀x(x = A →
φ))) |
12 | 10, 11 | syl6bb 252 |
. 2
⊢ ((Ⅎxψ ∧ ∀x(x = A → (φ
↔ ψ)) ∧ A ∈ B) →
(∀x
∈ B
(x = A
→ φ) ↔ (A ∈ B → ∀x(x = A →
φ)))) |
13 | | biimt 325 |
. . 3
⊢ (A ∈ B → (∀x(x = A →
φ) ↔ (A ∈ B → ∀x(x = A →
φ)))) |
14 | 13 | 3ad2ant3 978 |
. 2
⊢ ((Ⅎxψ ∧ ∀x(x = A → (φ
↔ ψ)) ∧ A ∈ B) →
(∀x(x = A → φ)
↔ (A ∈ B →
∀x(x = A → φ)))) |
15 | | ceqsalt 2882 |
. 2
⊢ ((Ⅎxψ ∧ ∀x(x = A → (φ
↔ ψ)) ∧ A ∈ B) →
(∀x(x = A → φ)
↔ ψ)) |
16 | 12, 14, 15 | 3bitr2d 272 |
1
⊢ ((Ⅎxψ ∧ ∀x(x = A → (φ
↔ ψ)) ∧ A ∈ B) →
(∀x
∈ B
(x = A
→ φ) ↔ ψ)) |