Users' Mathboxes Mathbox for Thierry Arnoux < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  indifbi Structured version   Visualization version   GIF version

Theorem indifbi 32606
Description: Two ways to express equality relative to a class 𝐴. (Contributed by Thierry Arnoux, 23-Jun-2024.)
Assertion
Ref Expression
indifbi ((𝐴𝐵) = (𝐴𝐶) ↔ (𝐴𝐵) = (𝐴𝐶))

Proof of Theorem indifbi
StepHypRef Expression
1 inss1 4191 . . 3 (𝐴𝐵) ⊆ 𝐴
2 inss1 4191 . . 3 (𝐴𝐶) ⊆ 𝐴
3 rcompleq 4259 . . 3 (((𝐴𝐵) ⊆ 𝐴 ∧ (𝐴𝐶) ⊆ 𝐴) → ((𝐴𝐵) = (𝐴𝐶) ↔ (𝐴 ∖ (𝐴𝐵)) = (𝐴 ∖ (𝐴𝐶))))
41, 2, 3mp2an 693 . 2 ((𝐴𝐵) = (𝐴𝐶) ↔ (𝐴 ∖ (𝐴𝐵)) = (𝐴 ∖ (𝐴𝐶)))
5 difin 4226 . . 3 (𝐴 ∖ (𝐴𝐵)) = (𝐴𝐵)
6 difin 4226 . . 3 (𝐴 ∖ (𝐴𝐶)) = (𝐴𝐶)
75, 6eqeq12i 2755 . 2 ((𝐴 ∖ (𝐴𝐵)) = (𝐴 ∖ (𝐴𝐶)) ↔ (𝐴𝐵) = (𝐴𝐶))
84, 7bitri 275 1 ((𝐴𝐵) = (𝐴𝐶) ↔ (𝐴𝐵) = (𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wb 206   = wceq 1542  cdif 3900  cin 3902  wss 3903
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 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1089  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-rab 3402  df-v 3444  df-dif 3906  df-in 3910  df-ss 3920
This theorem is referenced by:  fressupp  32777
  Copyright terms: Public domain W3C validator