Theorem difdif2ss 3221

Description: Set difference with a set difference. In classical logic this would be equality rather than subset. (Contributed by Jim Kingdon, 27-Jul-2018.)

Assertion

Ref	Expression
difdif2ss	|- ( ( A \ B ) u. ( A i^i C ) ) C_ ( A \ ( B \ C ) )

Proof of Theorem difdif2ss

Step	Hyp	Ref	Expression
1		inssdif 3200	. . . 4 |- ( A i^i C ) C_ ( A \ ( _V \ C ) )
2		unss2 3143	. . . 4 |- ( ( A i^i C ) C_ ( A \ ( _V \ C ) ) -> ( ( A \ B ) u. ( A i^i C ) ) C_ ( ( A \ B ) u. ( A \ ( _V \ C ) ) ) )
3	1, 2	ax-mp 7	. . 3 |- ( ( A \ B ) u. ( A i^i C ) ) C_ ( ( A \ B ) u. ( A \ ( _V \ C ) ) )
4		difindiss 3218	. . 3 |- ( ( A \ B ) u. ( A \ ( _V \ C ) ) ) C_ ( A \ ( B i^i ( _V \ C ) ) )
5	3, 4	sstri 3008	. 2 |- ( ( A \ B ) u. ( A i^i C ) ) C_ ( A \ ( B i^i ( _V \ C ) ) )
6		invdif 3206	. . . 4 |- ( B i^i ( _V \ C ) ) = ( B \ C )
7	6	eqcomi 2085	. . 3 |- ( B \ C ) = ( B i^i ( _V \ C ) )
8	7	difeq2i 3087	. 2 |- ( A \ ( B \ C ) ) = ( A \ ( B i^i ( _V \ C ) ) )
9	5, 8	sseqtr4i 3032	1 |- ( ( A \ B ) u. ( A i^i C ) ) C_ ( A \ ( B \ C ) )

Syntax hints:   _Vcvv 2601   \ cdif 2970   u. cun 2971   i^i cin 2972   C_ wss 2973

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rab 2357  df-v 2603  df-dif 2975  df-un 2977  df-in 2979  df-ss 2986

This theorem is referenced by: (None)