Theorem ineqcom 4107

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 4106	. 2 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	eqeq1i 2765	1 ⊢ ((𝐴 ∩ 𝐵) = 𝐶 ↔ (𝐵 ∩ 𝐴) = 𝐶)

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1569   ∩ cin 4089

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-rab 3413  df-in 4091

This theorem is used by:  sseqin2  4120  disjr  4403  uneqdifeq  4447  fnunres2  6652  padct  33310