Mathbox for Richard Penner < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  rcompleq Structured version   Visualization version   GIF version

Theorem rcompleq 39153
 Description: Two subclasses are equal if and only if their relative complements are equal. Relativized version of compleq 3981. (Contributed by RP, 10-Jun-2021.)
Assertion
Ref Expression
rcompleq ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))

Proof of Theorem rcompleq
StepHypRef Expression
1 ancom 454 . . 3 ((𝐴𝐵𝐵𝐴) ↔ (𝐵𝐴𝐴𝐵))
2 sscon34b 39152 . . . . 5 ((𝐵𝐶𝐴𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
32ancoms 452 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐵𝐴 ↔ (𝐶𝐴) ⊆ (𝐶𝐵)))
4 sscon34b 39152 . . . 4 ((𝐴𝐶𝐵𝐶) → (𝐴𝐵 ↔ (𝐶𝐵) ⊆ (𝐶𝐴)))
53, 4anbi12d 624 . . 3 ((𝐴𝐶𝐵𝐶) → ((𝐵𝐴𝐴𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
61, 5syl5bb 275 . 2 ((𝐴𝐶𝐵𝐶) → ((𝐴𝐵𝐵𝐴) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴))))
7 eqss 3842 . 2 (𝐴 = 𝐵 ↔ (𝐴𝐵𝐵𝐴))
8 eqss 3842 . 2 ((𝐶𝐴) = (𝐶𝐵) ↔ ((𝐶𝐴) ⊆ (𝐶𝐵) ∧ (𝐶𝐵) ⊆ (𝐶𝐴)))
96, 7, 83bitr4g 306 1 ((𝐴𝐶𝐵𝐶) → (𝐴 = 𝐵 ↔ (𝐶𝐴) = (𝐶𝐵)))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 386   = wceq 1656   ∖ cdif 3795   ⊆ wss 3798 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908  ax-5 2009  ax-6 2075  ax-7 2112  ax-9 2173  ax-10 2192  ax-11 2207  ax-12 2220  ax-ext 2803 This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-tru 1660  df-ex 1879  df-nf 1883  df-sb 2068  df-clab 2812  df-cleq 2818  df-clel 2821  df-nfc 2958  df-v 3416  df-dif 3801  df-in 3805  df-ss 3812 This theorem is referenced by:  ntrclsfveq1  39193  ntrclsfveq2  39194  ntrclskb  39202  ntrclsk13  39204  ntrclsk4  39205
 Copyright terms: Public domain W3C validator