Proof of Theorem dveeq2
Step | Hyp | Ref
| Expression |
1 | | ax12or 1501 |
. . . . 5
⊢
(∀𝑥 𝑥 = 𝑧 ∨ (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦))) |
2 | | orcom 723 |
. . . . . 6
⊢
((∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) ↔ (∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦) ∨ ∀𝑥 𝑥 = 𝑦)) |
3 | 2 | orbi2i 757 |
. . . . 5
⊢
((∀𝑥 𝑥 = 𝑧 ∨ (∀𝑥 𝑥 = 𝑦 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦))) ↔ (∀𝑥 𝑥 = 𝑧 ∨ (∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦) ∨ ∀𝑥 𝑥 = 𝑦))) |
4 | 1, 3 | mpbi 144 |
. . . 4
⊢
(∀𝑥 𝑥 = 𝑧 ∨ (∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦) ∨ ∀𝑥 𝑥 = 𝑦)) |
5 | | orass 762 |
. . . 4
⊢
(((∀𝑥 𝑥 = 𝑧 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) ∨ ∀𝑥 𝑥 = 𝑦) ↔ (∀𝑥 𝑥 = 𝑧 ∨ (∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦) ∨ ∀𝑥 𝑥 = 𝑦))) |
6 | 4, 5 | mpbir 145 |
. . 3
⊢
((∀𝑥 𝑥 = 𝑧 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) ∨ ∀𝑥 𝑥 = 𝑦) |
7 | | orel2 721 |
. . 3
⊢ (¬
∀𝑥 𝑥 = 𝑦 → (((∀𝑥 𝑥 = 𝑧 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) ∨ ∀𝑥 𝑥 = 𝑦) → (∀𝑥 𝑥 = 𝑧 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)))) |
8 | 6, 7 | mpi 15 |
. 2
⊢ (¬
∀𝑥 𝑥 = 𝑦 → (∀𝑥 𝑥 = 𝑧 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦))) |
9 | | ax16 1806 |
. . 3
⊢
(∀𝑥 𝑥 = 𝑧 → (𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) |
10 | | sp 1504 |
. . 3
⊢
(∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦) → (𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) |
11 | 9, 10 | jaoi 711 |
. 2
⊢
((∀𝑥 𝑥 = 𝑧 ∨ ∀𝑥(𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) → (𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) |
12 | 8, 11 | syl 14 |
1
⊢ (¬
∀𝑥 𝑥 = 𝑦 → (𝑧 = 𝑦 → ∀𝑥 𝑧 = 𝑦)) |