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Theorem indifbi 30868
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 4162 . . 3 (𝐴𝐵) ⊆ 𝐴
2 inss1 4162 . . 3 (𝐴𝐶) ⊆ 𝐴
3 rcompleq 4229 . . 3 (((𝐴𝐵) ⊆ 𝐴 ∧ (𝐴𝐶) ⊆ 𝐴) → ((𝐴𝐵) = (𝐴𝐶) ↔ (𝐴 ∖ (𝐴𝐵)) = (𝐴 ∖ (𝐴𝐶))))
41, 2, 3mp2an 689 . 2 ((𝐴𝐵) = (𝐴𝐶) ↔ (𝐴 ∖ (𝐴𝐵)) = (𝐴 ∖ (𝐴𝐶)))
5 difin 4195 . . 3 (𝐴 ∖ (𝐴𝐵)) = (𝐴𝐵)
6 difin 4195 . . 3 (𝐴 ∖ (𝐴𝐶)) = (𝐴𝐶)
75, 6eqeq12i 2756 . 2 ((𝐴 ∖ (𝐴𝐵)) = (𝐴 ∖ (𝐴𝐶)) ↔ (𝐴𝐵) = (𝐴𝐶))
84, 7bitri 274 1 ((𝐴𝐵) = (𝐴𝐶) ↔ (𝐴𝐵) = (𝐴𝐶))
Colors of variables: wff setvar class
Syntax hints:  wb 205   = wceq 1539  cdif 3884  cin 3886  wss 3887
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1542  df-ex 1783  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-rab 3073  df-v 3434  df-dif 3890  df-in 3894  df-ss 3904
This theorem is referenced by:  fressupp  31022
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