Proof of Theorem unixpid
Step | Hyp | Ref
| Expression |
1 | | xpeq1 5602 |
. . . 4
⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = (∅ × 𝐴)) |
2 | | 0xp 5683 |
. . . 4
⊢ (∅
× 𝐴) =
∅ |
3 | 1, 2 | eqtrdi 2795 |
. . 3
⊢ (𝐴 = ∅ → (𝐴 × 𝐴) = ∅) |
4 | | unieq 4855 |
. . . . 5
⊢ ((𝐴 × 𝐴) = ∅ → ∪ (𝐴
× 𝐴) = ∪ ∅) |
5 | 4 | unieqd 4858 |
. . . 4
⊢ ((𝐴 × 𝐴) = ∅ → ∪ ∪ (𝐴 × 𝐴) = ∪ ∪ ∅) |
6 | | uni0 4874 |
. . . . . 6
⊢ ∪ ∅ = ∅ |
7 | 6 | unieqi 4857 |
. . . . 5
⊢ ∪ ∪ ∅ = ∪ ∅ |
8 | 7, 6 | eqtri 2767 |
. . . 4
⊢ ∪ ∪ ∅ =
∅ |
9 | | eqtr 2762 |
. . . . 5
⊢ ((∪ ∪ (𝐴 × 𝐴) = ∪ ∪ ∅ ∧ ∪ ∪ ∅ = ∅) → ∪
∪ (𝐴 × 𝐴) = ∅) |
10 | | eqtr 2762 |
. . . . . . 7
⊢ ((∪ ∪ (𝐴 × 𝐴) = ∅ ∧ ∅ = 𝐴) → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
11 | 10 | expcom 413 |
. . . . . 6
⊢ (∅
= 𝐴 → (∪ ∪ (𝐴 × 𝐴) = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴)) |
12 | 11 | eqcoms 2747 |
. . . . 5
⊢ (𝐴 = ∅ → (∪ ∪ (𝐴 × 𝐴) = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴)) |
13 | 9, 12 | syl5com 31 |
. . . 4
⊢ ((∪ ∪ (𝐴 × 𝐴) = ∪ ∪ ∅ ∧ ∪ ∪ ∅ = ∅) → (𝐴 = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴)) |
14 | 5, 8, 13 | sylancl 585 |
. . 3
⊢ ((𝐴 × 𝐴) = ∅ → (𝐴 = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴)) |
15 | 3, 14 | mpcom 38 |
. 2
⊢ (𝐴 = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
16 | | df-ne 2945 |
. . 3
⊢ (𝐴 ≠ ∅ ↔ ¬ 𝐴 = ∅) |
17 | | xpnz 6059 |
. . . 4
⊢ ((𝐴 ≠ ∅ ∧ 𝐴 ≠ ∅) ↔ (𝐴 × 𝐴) ≠ ∅) |
18 | | unixp 6182 |
. . . . 5
⊢ ((𝐴 × 𝐴) ≠ ∅ → ∪ ∪ (𝐴 × 𝐴) = (𝐴 ∪ 𝐴)) |
19 | | unidm 4090 |
. . . . 5
⊢ (𝐴 ∪ 𝐴) = 𝐴 |
20 | 18, 19 | eqtrdi 2795 |
. . . 4
⊢ ((𝐴 × 𝐴) ≠ ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
21 | 17, 20 | sylbi 216 |
. . 3
⊢ ((𝐴 ≠ ∅ ∧ 𝐴 ≠ ∅) → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
22 | 16, 16, 21 | sylancbr 600 |
. 2
⊢ (¬
𝐴 = ∅ → ∪ ∪ (𝐴 × 𝐴) = 𝐴) |
23 | 15, 22 | pm2.61i 182 |
1
⊢ ∪ ∪ (𝐴 × 𝐴) = 𝐴 |