Proof of Theorem preleqg
| Step | Hyp | Ref | Expression | 
|---|
| 1 |  | elneq 9639 | . . . . 5
⊢ (𝐴 ∈ 𝐵 → 𝐴 ≠ 𝐵) | 
| 2 | 1 | 3ad2ant1 1133 | . . . 4
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → 𝐴 ≠ 𝐵) | 
| 3 |  | preq12nebg 4862 | . . . 4
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐴 ≠ 𝐵) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))) | 
| 4 | 2, 3 | syld3an3 1410 | . . 3
⊢ ((𝐴 ∈ 𝐵 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝐷) → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))) | 
| 5 |  | eleq12 2830 | . . . . . . . . . 10
⊢ ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 ∈ 𝐵 ↔ 𝐷 ∈ 𝐶)) | 
| 6 |  | elnotel 9651 | . . . . . . . . . . 11
⊢ (𝐷 ∈ 𝐶 → ¬ 𝐶 ∈ 𝐷) | 
| 7 | 6 | pm2.21d 121 | . . . . . . . . . 10
⊢ (𝐷 ∈ 𝐶 → (𝐶 ∈ 𝐷 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))) | 
| 8 | 5, 7 | biimtrdi 253 | . . . . . . . . 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 1110 | . . . . . 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 858 | . . . 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 240 | . 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)) |