Proof of Theorem pr1eqbg
Step | Hyp | Ref
| Expression |
1 | | eqid 2610 |
. . . . 5
⊢ 𝐵 = 𝐵 |
2 | 1 | biantru 525 |
. . . 4
⊢ (𝐴 = 𝐶 ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐵)) |
3 | 2 | orbi2i 540 |
. . 3
⊢ (((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ∨ 𝐴 = 𝐶) ↔ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ∨ (𝐴 = 𝐶 ∧ 𝐵 = 𝐵))) |
4 | 3 | a1i 11 |
. 2
⊢ (((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) ∧ 𝐴 ≠ 𝐵) → (((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ∨ 𝐴 = 𝐶) ↔ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ∨ (𝐴 = 𝐶 ∧ 𝐵 = 𝐵)))) |
5 | | df-ne 2782 |
. . . . . 6
⊢ (𝐴 ≠ 𝐵 ↔ ¬ 𝐴 = 𝐵) |
6 | 5 | biimpi 205 |
. . . . 5
⊢ (𝐴 ≠ 𝐵 → ¬ 𝐴 = 𝐵) |
7 | 6 | adantl 481 |
. . . 4
⊢ (((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) ∧ 𝐴 ≠ 𝐵) → ¬ 𝐴 = 𝐵) |
8 | 7 | intnanrd 954 |
. . 3
⊢ (((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) ∧ 𝐴 ≠ 𝐵) → ¬ (𝐴 = 𝐵 ∧ 𝐵 = 𝐶)) |
9 | | biorf 419 |
. . 3
⊢ (¬
(𝐴 = 𝐵 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ↔ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ∨ 𝐴 = 𝐶))) |
10 | 8, 9 | syl 17 |
. 2
⊢ (((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) ∧ 𝐴 ≠ 𝐵) → (𝐴 = 𝐶 ↔ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ∨ 𝐴 = 𝐶))) |
11 | | 3simpa 1051 |
. . . . 5
⊢ ((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) → (𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉)) |
12 | | 3simpc 1053 |
. . . . 5
⊢ ((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) → (𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋)) |
13 | 11, 12 | jca 553 |
. . . 4
⊢ ((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) → ((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉) ∧ (𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋))) |
14 | 13 | adantr 480 |
. . 3
⊢ (((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) ∧ 𝐴 ≠ 𝐵) → ((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉) ∧ (𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋))) |
15 | | preq12bg 4326 |
. . 3
⊢ (((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉) ∧ (𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋)) → ({𝐴, 𝐵} = {𝐵, 𝐶} ↔ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ∨ (𝐴 = 𝐶 ∧ 𝐵 = 𝐵)))) |
16 | 14, 15 | syl 17 |
. 2
⊢ (((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) ∧ 𝐴 ≠ 𝐵) → ({𝐴, 𝐵} = {𝐵, 𝐶} ↔ ((𝐴 = 𝐵 ∧ 𝐵 = 𝐶) ∨ (𝐴 = 𝐶 ∧ 𝐵 = 𝐵)))) |
17 | 4, 10, 16 | 3bitr4d 299 |
1
⊢ (((𝐴 ∈ 𝑈 ∧ 𝐵 ∈ 𝑉 ∧ 𝐶 ∈ 𝑋) ∧ 𝐴 ≠ 𝐵) → (𝐴 = 𝐶 ↔ {𝐴, 𝐵} = {𝐵, 𝐶})) |