Theorem ineqcom 4110

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 4109	. 2 ⊢ (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐴)
2	1	eqeq1i 2768	1 ⊢ ((𝐴 ∩ 𝐵) = 𝐶 ↔ (𝐵 ∩ 𝐴) = 𝐶)

This proof depends on syntax axioms:   ↔ wb 209   = wceq 1570   ∩ 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-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-rab 3417  df-in 4094

This theorem is used by:  sseqin2  4123  disjr  4407  uneqdifeq  4451  fnunres2  6671  padct  33378