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Theorem pr1eqbg 4881
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 2740 . . . . 5 𝐵 = 𝐵
21biantru 529 . . . 4 (𝐴 = 𝐶 ↔ (𝐴 = 𝐶𝐵 = 𝐵))
32orbi2i 911 . . 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 2952 . . . . 5 (𝐴𝐵 → ¬ 𝐴 = 𝐵)
65adantl 481 . . . 4 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → ¬ 𝐴 = 𝐵)
76intnanrd 489 . . 3 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → ¬ (𝐴 = 𝐵𝐵 = 𝐶))
8 biorf 935 . . 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 1148 . . . . 5 ((𝐴𝑈𝐵𝑉𝐶𝑋) → (𝐴𝑈𝐵𝑉))
11 3simpc 1150 . . . . 5 ((𝐴𝑈𝐵𝑉𝐶𝑋) → (𝐵𝑉𝐶𝑋))
1210, 11jca 511 . . . 4 ((𝐴𝑈𝐵𝑉𝐶𝑋) → ((𝐴𝑈𝐵𝑉) ∧ (𝐵𝑉𝐶𝑋)))
1312adantr 480 . . 3 (((𝐴𝑈𝐵𝑉𝐶𝑋) ∧ 𝐴𝐵) → ((𝐴𝑈𝐵𝑉) ∧ (𝐵𝑉𝐶𝑋)))
14 preq12bg 4878 . . 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 206  wa 395  wo 846  w3a 1087   = wceq 1537  wcel 2108  wne 2946  {cpr 4650
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ne 2947  df-v 3490  df-un 3981  df-sn 4649  df-pr 4651
This theorem is referenced by:  pr1nebg  4882
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