| Step | Hyp | Ref
 | Expression | 
| 1 |   | xpeq12 4682 | 
. . . . 5
⊢ ((𝐴 = 𝐵 ∧ 𝐴 = 𝐵) → (𝐴 × 𝐴) = (𝐵 × 𝐵)) | 
| 2 | 1 | anidms 397 | 
. . . 4
⊢ (𝐴 = 𝐵 → (𝐴 × 𝐴) = (𝐵 × 𝐵)) | 
| 3 | 2 | sseq2d 3213 | 
. . 3
⊢ (𝐴 = 𝐵 → (𝑅 ⊆ (𝐴 × 𝐴) ↔ 𝑅 ⊆ (𝐵 × 𝐵))) | 
| 4 |   | raleq 2693 | 
. . . 4
⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥 ↔ ∀𝑥 ∈ 𝐵 ¬ 𝑥𝑅𝑥)) | 
| 5 |   | raleq 2693 | 
. . . . 5
⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥) ↔ ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥))) | 
| 6 | 5 | raleqbi1dv 2705 | 
. . . 4
⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥) ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥))) | 
| 7 | 4, 6 | anbi12d 473 | 
. . 3
⊢ (𝐴 = 𝐵 → ((∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ↔ (∀𝑥 ∈ 𝐵 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)))) | 
| 8 |   | raleq 2693 | 
. . . . . 6
⊢ (𝐴 = 𝐵 → (∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ↔ ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)))) | 
| 9 | 8 | raleqbi1dv 2705 | 
. . . . 5
⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ↔ ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)))) | 
| 10 | 9 | raleqbi1dv 2705 | 
. . . 4
⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)))) | 
| 11 |   | raleq 2693 | 
. . . . 5
⊢ (𝐴 = 𝐵 → (∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦) ↔ ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦))) | 
| 12 | 11 | raleqbi1dv 2705 | 
. . . 4
⊢ (𝐴 = 𝐵 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦) ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦))) | 
| 13 | 10, 12 | anbi12d 473 | 
. . 3
⊢ (𝐴 = 𝐵 → ((∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦)) ↔ (∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦)))) | 
| 14 | 3, 7, 13 | 3anbi123d 1323 | 
. 2
⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ (𝐴 × 𝐴) ∧ (∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ∧ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦))) ↔ (𝑅 ⊆ (𝐵 × 𝐵) ∧ (∀𝑥 ∈ 𝐵 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ∧ (∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦))))) | 
| 15 |   | dftap2 7318 | 
. 2
⊢ (𝑅 TAp 𝐴 ↔ (𝑅 ⊆ (𝐴 × 𝐴) ∧ (∀𝑥 ∈ 𝐴 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ∧ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦)))) | 
| 16 |   | dftap2 7318 | 
. 2
⊢ (𝑅 TAp 𝐵 ↔ (𝑅 ⊆ (𝐵 × 𝐵) ∧ (∀𝑥 ∈ 𝐵 ¬ 𝑥𝑅𝑥 ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (𝑥𝑅𝑦 → 𝑦𝑅𝑥)) ∧ (∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (𝑥𝑅𝑦 → (𝑥𝑅𝑧 ∨ 𝑦𝑅𝑧)) ∧ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 (¬ 𝑥𝑅𝑦 → 𝑥 = 𝑦)))) | 
| 17 | 14, 15, 16 | 3bitr4g 223 | 
1
⊢ (𝐴 = 𝐵 → (𝑅 TAp 𝐴 ↔ 𝑅 TAp 𝐵)) |