Proof of Theorem eupth2lem1
Step | Hyp | Ref
| Expression |
1 | | eleq2 2828 |
. . 3
⊢ (∅
= if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) → (𝑈 ∈ ∅ ↔ 𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}))) |
2 | 1 | bibi1d 343 |
. 2
⊢ (∅
= if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) → ((𝑈 ∈ ∅ ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵))) ↔ (𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵))))) |
3 | | eleq2 2828 |
. . 3
⊢ ({𝐴, 𝐵} = if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) → (𝑈 ∈ {𝐴, 𝐵} ↔ 𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}))) |
4 | 3 | bibi1d 343 |
. 2
⊢ ({𝐴, 𝐵} = if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) → ((𝑈 ∈ {𝐴, 𝐵} ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵))) ↔ (𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵))))) |
5 | | noel 4269 |
. . . 4
⊢ ¬
𝑈 ∈
∅ |
6 | 5 | a1i 11 |
. . 3
⊢ ((𝑈 ∈ 𝑉 ∧ 𝐴 = 𝐵) → ¬ 𝑈 ∈ ∅) |
7 | | simpl 482 |
. . . . 5
⊢ ((𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵)) → 𝐴 ≠ 𝐵) |
8 | 7 | neneqd 2949 |
. . . 4
⊢ ((𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵)) → ¬ 𝐴 = 𝐵) |
9 | | simpr 484 |
. . . 4
⊢ ((𝑈 ∈ 𝑉 ∧ 𝐴 = 𝐵) → 𝐴 = 𝐵) |
10 | 8, 9 | nsyl3 138 |
. . 3
⊢ ((𝑈 ∈ 𝑉 ∧ 𝐴 = 𝐵) → ¬ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵))) |
11 | 6, 10 | 2falsed 376 |
. 2
⊢ ((𝑈 ∈ 𝑉 ∧ 𝐴 = 𝐵) → (𝑈 ∈ ∅ ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵)))) |
12 | | elprg 4587 |
. . 3
⊢ (𝑈 ∈ 𝑉 → (𝑈 ∈ {𝐴, 𝐵} ↔ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵))) |
13 | | df-ne 2945 |
. . . 4
⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵) |
14 | | ibar 528 |
. . . 4
⊢ (𝐴 ≠ 𝐵 → ((𝑈 = 𝐴 ∨ 𝑈 = 𝐵) ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵)))) |
15 | 13, 14 | sylbir 234 |
. . 3
⊢ (¬
𝐴 = 𝐵 → ((𝑈 = 𝐴 ∨ 𝑈 = 𝐵) ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵)))) |
16 | 12, 15 | sylan9bb 509 |
. 2
⊢ ((𝑈 ∈ 𝑉 ∧ ¬ 𝐴 = 𝐵) → (𝑈 ∈ {𝐴, 𝐵} ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵)))) |
17 | 2, 4, 11, 16 | ifbothda 4502 |
1
⊢ (𝑈 ∈ 𝑉 → (𝑈 ∈ if(𝐴 = 𝐵, ∅, {𝐴, 𝐵}) ↔ (𝐴 ≠ 𝐵 ∧ (𝑈 = 𝐴 ∨ 𝑈 = 𝐵)))) |