Theorem difdif2ss 3257

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 3236	. . . 4 |- ( A i^i C ) C_ ( A \ ( _V \ C ) )
2		unss2 3172	. . . 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 3254	. . 3 |- ( ( A \ B ) u. ( A \ ( _V \ C ) ) ) C_ ( A \ ( B i^i ( _V \ C ) ) )
5	3, 4	sstri 3035	. 2 |- ( ( A \ B ) u. ( A i^i C ) ) C_ ( A \ ( B i^i ( _V \ C ) ) )
6		invdif 3242	. . . 4 |- ( B i^i ( _V \ C ) ) = ( B \ C )
7	6	eqcomi 2093	. . 3 |- ( B \ C ) = ( B i^i ( _V \ C ) )
8	7	difeq2i 3116	. 2 |- ( A \ ( B \ C ) ) = ( A \ ( B i^i ( _V \ C ) ) )
9	5, 8	sseqtr4i 3060	1 |- ( ( A \ B ) u. ( A i^i C ) ) C_ ( A \ ( B \ C ) )

Syntax hints:   _Vcvv 2620   \ cdif 2997   u. cun 2998   i^i cin 2999   C_ wss 3000

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 580  ax-in2 581  ax-io 666  ax-5 1382  ax-7 1383  ax-gen 1384  ax-ie1 1428  ax-ie2 1429  ax-8 1441  ax-10 1442  ax-11 1443  ax-i12 1444  ax-bndl 1445  ax-4 1446  ax-17 1465  ax-i9 1469  ax-ial 1473  ax-i5r 1474  ax-ext 2071

This theorem depends on definitions:  df-bi 116  df-tru 1293  df-nf 1396  df-sb 1694  df-clab 2076  df-cleq 2082  df-clel 2085  df-nfc 2218  df-ral 2365  df-rab 2369  df-v 2622  df-dif 3002  df-un 3004  df-in 3006  df-ss 3013

This theorem is referenced by: (None)