Theorem indifbi 33191

Description: Two ways to express equality relative to a class 𝐴. (Contributed by Thierry Arnoux, 23-Jun-2024.)

Assertion

Ref	Expression
indifbi	⊢ ((𝐴 ∩ 𝐵) = (𝐴 ∩ 𝐶) ↔ (𝐴 ∖ 𝐵) = (𝐴 ∖ 𝐶))

Proof of Theorem indifbi

Step	Hyp	Ref	Expression
1		inss1 4136	. . 3 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴
2		inss1 4136	. . 3 ⊢ (𝐴 ∩ 𝐶) ⊆ 𝐴
3		rcompleq 4253	. . 3 ⊢ (((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐶) ⊆ 𝐴) → ((𝐴 ∩ 𝐵) = (𝐴 ∩ 𝐶) ↔ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ (𝐴 ∩ 𝐶))))
4	1, 2, 3	mp2an 705	. 2 ⊢ ((𝐴 ∩ 𝐵) = (𝐴 ∩ 𝐶) ↔ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ (𝐴 ∩ 𝐶)))
5		difin 4220	. . 3 ⊢ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ 𝐵)
6		difin 4220	. . 3 ⊢ (𝐴 ∖ (𝐴 ∩ 𝐶)) = (𝐴 ∖ 𝐶)
7	5, 6	eqeq12i 2781	. 2 ⊢ ((𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ (𝐴 ∩ 𝐶)) ↔ (𝐴 ∖ 𝐵) = (𝐴 ∖ 𝐶))
8	4, 7	bitri 278	1 ⊢ ((𝐴 ∩ 𝐵) = (𝐴 ∩ 𝐶) ↔ (𝐴 ∖ 𝐵) = (𝐴 ∖ 𝐶))

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570   ⊆ wss 3900   ∖ cdif 4048   ∩ cin 4092

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094

This theorem is used by:  fressupp  33355