Proof of Theorem unixpid
| Step | Hyp | Ref
| Expression |
| 1 | | xpeq1 5681 |
. . . 4
⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × 𝐴)) |
| 2 | | 0xp 5766 |
. . . 4
⊢ (∅
× 𝐴) =
∅ |
| 3 | 1, 2 | eqtrdi 2785 |
. . 3
⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = ∅) |
| 4 | | unieq 4900 |
. . . . 5
⊢ ((𝐴 × 𝐴) = ∅ → ∪ (𝐴
× 𝐴) = ∪ ∅) |
| 5 | 4 | unieqd 4902 |
. . . 4
⊢ ((𝐴 × 𝐴) = ∅ → ∪ ∪ (𝐴 × 𝐴) = ∪ ∪ ∅) |
| 6 | | uni0 4917 |
. . . . . 6
⊢ ∪ ∅ = ∅ |
| 7 | 6 | unieqi 4901 |
. . . . 5
⊢ ∪ ∪ ∅ = ∪ ∅ |
| 8 | 7, 6 | eqtri 2757 |
. . . 4
⊢ ∪ ∪ ∅ =
∅ |
| 9 | | eqtr 2754 |
. . . . 5
⊢ ((∪ ∪ (𝐴 × 𝐴) = ∪ ∪ ∅ ∧ ∪ ∪ ∅ = ∅) → ∪
∪ (𝐴 × 𝐴) = ∅) |
| 10 | | eqtr 2754 |
. . . . . . 7
⊢ ((∪ ∪ (𝐴 × 𝐴) = ∅ ∧ ∅ = 𝐴) → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
| 11 | 10 | expcom 413 |
. . . . . 6
⊢ (∅
= 𝐴 → (∪ ∪ (𝐴 × 𝐴) = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴)) |
| 12 | 11 | eqcoms 2742 |
. . . . 5
⊢ (𝐴 = ∅ → (∪ ∪ (𝐴 × 𝐴) = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴)) |
| 13 | 9, 12 | syl5com 31 |
. . . 4
⊢ ((∪ ∪ (𝐴 × 𝐴) = ∪ ∪ ∅ ∧ ∪ ∪ ∅ = ∅) → (𝐴 = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴)) |
| 14 | 5, 8, 13 | sylancl 586 |
. . 3
⊢ ((𝐴 × 𝐴) = ∅ → (𝐴 = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴)) |
| 15 | 3, 14 | mpcom 38 |
. 2
⊢ (𝐴 = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
| 16 | | df-ne 2932 |
. . 3
⊢ (𝐴 ≠ ∅ ↔ ¬ 𝐴 = ∅) |
| 17 | | xpnz 6161 |
. . . 4
⊢ ((𝐴 ≠ ∅ ∧ 𝐴 ≠ ∅) ↔ (𝐴 × 𝐴) ≠ ∅) |
| 18 | | unixp 6284 |
. . . . 5
⊢ ((𝐴 × 𝐴) ≠ ∅ → ∪ ∪ (𝐴 × 𝐴) = (𝐴 ∪ 𝐴)) |
| 19 | | unidm 4139 |
. . . . 5
⊢ (𝐴 ∪ 𝐴) = 𝐴 |
| 20 | 18, 19 | eqtrdi 2785 |
. . . 4
⊢ ((𝐴 × 𝐴) ≠ ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
| 21 | 17, 20 | sylbi 217 |
. . 3
⊢ ((𝐴 ≠ ∅ ∧ 𝐴 ≠ ∅) → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
| 22 | 16, 16, 21 | sylancbr 601 |
. 2
⊢ (¬
𝐴 = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
| 23 | 15, 22 | pm2.61i 182 |
1
⊢ ∪ ∪ (𝐴 × 𝐴) = 𝐴 |