Theorem indifbi 32502

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 4186	. . 3 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴
2		inss1 4186	. . 3 ⊢ (𝐴 ∩ 𝐶) ⊆ 𝐴
3		rcompleq 4254	. . 3 ⊢ (((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐶) ⊆ 𝐴) → ((𝐴 ∩ 𝐵) = (𝐴 ∩ 𝐶) ↔ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ (𝐴 ∩ 𝐶))))
4	1, 2, 3	mp2an 692	. 2 ⊢ ((𝐴 ∩ 𝐵) = (𝐴 ∩ 𝐶) ↔ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ (𝐴 ∩ 𝐶)))
5		difin 4221	. . 3 ⊢ (𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ 𝐵)
6		difin 4221	. . 3 ⊢ (𝐴 ∖ (𝐴 ∩ 𝐶)) = (𝐴 ∖ 𝐶)
7	5, 6	eqeq12i 2751	. 2 ⊢ ((𝐴 ∖ (𝐴 ∩ 𝐵)) = (𝐴 ∖ (𝐴 ∩ 𝐶)) ↔ (𝐴 ∖ 𝐵) = (𝐴 ∖ 𝐶))
8	4, 7	bitri 275	1 ⊢ ((𝐴 ∩ 𝐵) = (𝐴 ∩ 𝐶) ↔ (𝐴 ∖ 𝐵) = (𝐴 ∖ 𝐶))

Syntax hints:   ↔ wb 206   = wceq 1541   ∖ cdif 3895   ∩ cin 3897   ⊆ wss 3898

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-ext 2705

This theorem depends on definitions:  df-bi 207  df-an 396  df-3an 1088  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2712  df-cleq 2725  df-clel 2808  df-rab 3397  df-v 3439  df-dif 3901  df-in 3905  df-ss 3915

This theorem is referenced by:  fressupp  32673