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Theorem pr1eqbg 4798
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 2736 . . . . 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 2946 . . . . 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 4795 . . 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 310 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 1540  wcel 2105  wne 2940  {cpr 4572
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010  ax-8 2107  ax-9 2115  ax-ext 2707
This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1543  df-ex 1781  df-sb 2067  df-clab 2714  df-cleq 2728  df-clel 2814  df-ne 2941  df-v 3442  df-un 3901  df-sn 4571  df-pr 4573
This theorem is referenced by:  pr1nebg  4799
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