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Theorem pr1eqbg 4787
Description: A (proper) pair is equal to another (maybe improper) pair containing one element of the first pair if and only if the other element of the first pair is contained in the second pair. (Contributed by Alexander van der Vekens, 26-Jan-2018.)
Assertion
Ref Expression
pr1eqbg (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → (𝐴 = 𝐶 ↔ {𝐴, 𝐵} = {𝐵, 𝐶}))

Proof of Theorem pr1eqbg
StepHypRef Expression
1 eqid 2738 . . . . 5 𝐵 = 𝐵
21biantru 530 . . . 4 (𝐴 = 𝐶 ↔ (𝐴 = 𝐶𝐵 = 𝐵))
32orbi2i 910 . . 3 (((𝐴 = 𝐵𝐵 = 𝐶) ∨ 𝐴 = 𝐶) ↔ ((𝐴 = 𝐵𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐵)))
43a1i 11 . 2 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → (((𝐴 = 𝐵𝐵 = 𝐶) ∨ 𝐴 = 𝐶) ↔ ((𝐴 = 𝐵𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐵))))
5 neneq 2949 . . . . 5 (𝐴𝐵 → ¬ 𝐴 = 𝐵)
65adantl 482 . . . 4 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → ¬ 𝐴 = 𝐵)
76intnanrd 490 . . 3 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → ¬ (𝐴 = 𝐵𝐵 = 𝐶))
8 biorf 934 . . 3 (¬ (𝐴 = 𝐵𝐵 = 𝐶) → (𝐴 = 𝐶 ↔ ((𝐴 = 𝐵𝐵 = 𝐶) ∨ 𝐴 = 𝐶)))
97, 8syl 17 . 2 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → (𝐴 = 𝐶 ↔ ((𝐴 = 𝐵𝐵 = 𝐶) ∨ 𝐴 = 𝐶)))
10 3simpa 1147 . . . . 5 ((𝐴𝑈𝐵𝑉𝐶𝑋) → (𝐴𝑈𝐵𝑉))
11 3simpc 1149 . . . . 5 ((𝐴𝑈𝐵𝑉𝐶𝑋) → (𝐵𝑉𝐶𝑋))
1210, 11jca 512 . . . 4 ((𝐴𝑈𝐵𝑉𝐶𝑋) → ((𝐴𝑈𝐵𝑉) ∧ (𝐵𝑉𝐶𝑋)))
1312adantr 481 . . 3 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → ((𝐴𝑈𝐵𝑉) ∧ (𝐵𝑉𝐶𝑋)))
14 preq12bg 4784 . . 3 (((𝐴𝑈𝐵𝑉) ∧ (𝐵𝑉𝐶𝑋)) → ({𝐴, 𝐵} = {𝐵, 𝐶} ↔ ((𝐴 = 𝐵𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐵))))
1513, 14syl 17 . 2 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → ({𝐴, 𝐵} = {𝐵, 𝐶} ↔ ((𝐴 = 𝐵𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐵))))
164, 9, 153bitr4d 311 1 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → (𝐴 = 𝐶 ↔ {𝐴, 𝐵} = {𝐵, 𝐶}))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 205  wa 396  wo 844  w3a 1086   = wceq 1539  wcel 2106  wne 2943  {cpr 4563
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ne 2944  df-v 3434  df-un 3892  df-sn 4562  df-pr 4564
This theorem is referenced by:  pr1nebg  4788
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