Theorem ineqcom 3422

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 3421	. 2 |- ( A i^i B ) = ( B i^i A )
2	1	eqeq1i 2246	1 |- ( ( A i^i B ) = C <-> ( B i^i A ) = C )

Syntax hints:   <-> wb 105   = wceq 1402   i^i cin 3219

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 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-in 3226

This theorem is referenced by:  ballotfilemfp1  13214