Theorem ineqcom 3411

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 i^i B ) = (/) <-> ( B i^i A ) = (/) )). (Contributed by Peter Mazsa, 22-Mar-2017.)

Assertion

Ref	Expression
ineqcom	|- ( ( A i^i B ) = C <-> ( B i^i A ) = C )

Proof of Theorem ineqcom

Step	Hyp	Ref	Expression
1		incom 3410	. 2 |- ( A i^i B ) = ( B i^i A )
2	1	eqeq1i 2240	1 |- ( ( A i^i B ) = C <-> ( B i^i A ) = C )

Syntax hints:   <-> wb 105   = wceq 1398   i^i cin 3209

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-v 2814  df-in 3216

This theorem is referenced by: (None)