Proof of Theorem sbco2
| Step | Hyp | Ref
| Expression |
| 1 | | sbequ 877 |
. . . . 5
⊢ (x =
y → ([x / z][z / x]φ ↔ [y / z][z / x]φ)) |
| 2 | | sbco2.1 |
. . . . . 6
⊢ (φ
→ ∀zφ) |
| 3 | 2 | sbid2 911 |
. . . . 5
⊢ ([x /
z][z /
x]φ
↔ φ) |
| 4 | 1, 3 | syl5bbr 412 |
. . . 4
⊢ (x =
y → (φ ↔ [y / z][z / x]φ)) |
| 5 | | sbequ12 865 |
. . . 4
⊢ (x =
y → (φ ↔ [y / x]φ)) |
| 6 | 4, 5 | bitr3d 408 |
. . 3
⊢ (x =
y → ([y / z][z / x]φ ↔ [y / x]φ)) |
| 7 | 6 | a4s 682 |
. 2
⊢ (∀x x = y → ([y /
z][z /
x]φ
↔ [y / x]φ)) |
| 8 | | eq6 826 |
. . . 4
⊢ (¬ ∀x x = y → ∀x ¬ ∀x x = y) |
| 9 | 2 | hbsb3 875 |
. . . . 5
⊢ ([z /
x]φ
→ ∀x[z / x]φ) |
| 10 | 9 | hbsb4 905 |
. . . 4
⊢ (¬ ∀x x = y → ([y /
z][z /
x]φ
→ ∀x[y / z][z / x]φ)) |
| 11 | 4 | a1i 7 |
. . . 4
⊢ (¬ ∀x x = y → (x =
y → (φ ↔ [y / z][z / x]φ))) |
| 12 | 8, 10, 11 | sbied 903 |
. . 3
⊢ (¬ ∀x x = y → ([y /
x]φ
↔ [y / z][z / x]φ)) |
| 13 | 12 | bicomd 399 |
. 2
⊢ (¬ ∀x x = y → ([y /
z][z /
x]φ
↔ [y / x]φ)) |
| 14 | 7, 13 | pm2.61i 110 |
1
⊢ ([y /
z][z /
x]φ
↔ [y / x]φ) |