Theorem ineqcom 4162

Description: Two ways of expressing that two classes have a given intersection. This is often used when that given intersection is the empty set, in which case the statement displays two ways of expressing that two classes are disjoint (when 𝐶 = ∅: ((𝐴 ∩ 𝐵) = ∅ ↔ (𝐵 ∩ 𝐴) = ∅)). (Contributed by Peter Mazsa, 22-Mar-2017.)

Assertion

Ref	Expression
ineqcom	⊢ ((𝐴 ∩ 𝐵) = 𝐶 ↔ (𝐵 ∩ 𝐴) = 𝐶)

Proof of Theorem ineqcom

Step	Hyp	Ref	Expression
1		incom 4161	. 2 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	eqeq1i 2767	1 ⊢ ((𝐴 ∩ 𝐵) = 𝐶 ↔ (𝐵 ∩ 𝐴) = 𝐶)

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1569   ∩ cin 3903

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-rab 3416  df-in 3911

This theorem is used by:  sseqin2  4175  disjr  4410  uneqdifeq  4452  fnunres2  6648  padct  33074