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Theorem rcompleq 40670
 Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 4110. (Contributed by RP, 10-Jun-2021.)
Assertion
Ref Expression
rcompleq ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))

Proof of Theorem rcompleq
StepHypRef Expression
1 ancom 464 . . 3 ((𝐴𝐵𝐵𝐴) ↔ (𝐵𝐴𝐴𝐵))
2 sscon34b 40669 . . . . 5 ((𝐵𝐶𝐴𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
32ancoms 462 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
4 sscon34b 40669 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
53, 4anbi12d 633 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐵𝐴𝐴𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
61, 5syl5bb 286 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐴𝐵𝐵𝐴) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
7 eqss 3968 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
8 eqss 3968 . 2 ((𝐶𝐴) = (𝐶𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴)))
96, 7, 83bitr4g 317 1 ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   = wceq 1538   ∖ cdif 3916   ⊆ wss 3919 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1971  ax-7 2016  ax-8 2117  ax-9 2125  ax-ext 2796 This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1541  df-ex 1782  df-sb 2071  df-clab 2803  df-cleq 2817  df-clel 2896  df-rab 3142  df-v 3482  df-dif 3922  df-in 3926  df-ss 3936 This theorem is referenced by:  ntrclsfveq1  40710  ntrclsfveq2  40711  ntrclskb  40719  ntrclsk13  40721  ntrclsk4  40722
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