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Theorem ineqcom 4117
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 = ∅: ((𝐴𝐵) = ∅ ↔ (𝐵𝐴) = ∅)). (Contributed by Peter Mazsa, 22-Mar-2017.)
Assertion
Ref Expression
ineqcom ((𝐴𝐵) = 𝐶 ↔ (𝐵𝐴) = 𝐶)

Proof of Theorem ineqcom
StepHypRef Expression
1 incom 4115 . 2 (𝐴𝐵) = (𝐵𝐴)
21eqeq1i 2742 1 ((𝐴𝐵) = 𝐶 ↔ (𝐵𝐴) = 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wb 209   = wceq 1543  cin 3865
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-9 2120  ax-ext 2708
This theorem depends on definitions:  df-bi 210  df-an 400  df-tru 1546  df-ex 1788  df-sb 2071  df-clab 2715  df-cleq 2729  df-rab 3070  df-in 3873
This theorem is referenced by: (None)
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