Proof of Theorem preleqg
Step | Hyp | Ref
| Expression |
1 | | elneq 9287 |
. . . . 5
⊢ (𝐴 ∈ 𝐵 → 𝐴 ≠ 𝐵) |
2 | 1 | 3ad2ant1 1131 |
. . . 4
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → 𝐴 ≠ 𝐵) |
3 | | preq12nebg 4790 |
. . . 4
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐴 ≠ 𝐵) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))) |
4 | 2, 3 | syld3an3 1407 |
. . 3
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))) |
5 | | eleq12 2828 |
. . . . . . . . . 10
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 ∈ 𝐵 ↔ 𝐷 ∈ 𝐶)) |
6 | | elnotel 9298 |
. . . . . . . . . . 11
⊢ (𝐷 ∈ 𝐶 → ¬ 𝐶 ∈ 𝐷) |
7 | 6 | pm2.21d 121 |
. . . . . . . . . 10
⊢ (𝐷 ∈ 𝐶 → (𝐶 ∈ 𝐷 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) |
8 | 5, 7 | syl6bi 252 |
. . . . . . . . 9
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 ∈ 𝐵 → (𝐶 ∈ 𝐷 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))) |
9 | 8 | com3l 89 |
. . . . . . . 8
⊢ (𝐴 ∈ 𝐵 → (𝐶 ∈ 𝐷 → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))) |
10 | 9 | a1d 25 |
. . . . . . 7
⊢ (𝐴 ∈ 𝐵 → (𝐵 ∈ 𝑉 → (𝐶 ∈ 𝐷 → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))) |
11 | 10 | 3imp 1109 |
. . . . . 6
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) |
12 | 11 | com12 32 |
. . . . 5
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) |
13 | 12 | jao1i 854 |
. . . 4
⊢ (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) → ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) |
14 | 13 | com12 32 |
. . 3
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) |
15 | 4, 14 | sylbid 239 |
. 2
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → ({𝐴, 𝐵} = {𝐶, 𝐷} → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) |
16 | 15 | imp 406 |
1
⊢ (((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) ∧ {𝐴, 𝐵} = {𝐶, 𝐷}) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)) |