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Theorem ineqcom 4197
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 4196 . 2 (𝐴𝐵) = (𝐵𝐴)
21eqeq1i 2731 1 ((𝐴𝐵) = 𝐶 ↔ (𝐵𝐴) = 𝐶)
Colors of variables: wff setvar class
Syntax hints:  wb 205   = wceq 1533  cin 3942
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-9 2108  ax-ext 2697
This theorem depends on definitions:  df-bi 206  df-an 396  df-tru 1536  df-ex 1774  df-sb 2060  df-clab 2704  df-cleq 2718  df-rab 3427  df-in 3950
This theorem is referenced by:  fnunres2  6655
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