Theorem pssdifcom1 4054

Description: Two ways to express overlapping subsets. (Contributed by Stefan O'Rear, 31-Oct-2014.)

Assertion

Ref	Expression
pssdifcom1	|- ( ( A C_ C /\ B C_ C ) -> ( ( C \ A ) C. B <-> ( C \ B ) C. A ) )

Proof of Theorem pssdifcom1

Step	Hyp	Ref	Expression
1		difcom 4053	. . . 4 |- ( ( C \ A ) C_ B <-> ( C \ B ) C_ A )
2	1	a1i 11	. . 3 |- ( ( A C_ C /\ B C_ C ) -> ( ( C \ A ) C_ B <-> ( C \ B ) C_ A ) )
3		ssconb 3743	. . . . 5 |- ( ( B C_ C /\ A C_ C ) -> ( B C_ ( C \ A ) <-> A C_ ( C \ B ) ) )
4	3	ancoms 469	. . . 4 |- ( ( A C_ C /\ B C_ C ) -> ( B C_ ( C \ A ) <-> A C_ ( C \ B ) ) )
5	4	notbid 308	. . 3 |- ( ( A C_ C /\ B C_ C ) -> ( -. B C_ ( C \ A ) <-> -. A C_ ( C \ B ) ) )
6	2, 5	anbi12d 747	. 2 |- ( ( A C_ C /\ B C_ C ) -> ( ( ( C \ A ) C_ B /\ -. B C_ ( C \ A ) ) <-> ( ( C \ B ) C_ A /\ -. A C_ ( C \ B ) ) ) )
7		dfpss3 3693	. 2 |- ( ( C \ A ) C. B <-> ( ( C \ A ) C_ B /\ -. B C_ ( C \ A ) ) )
8		dfpss3 3693	. 2 |- ( ( C \ B ) C. A <-> ( ( C \ B ) C_ A /\ -. A C_ ( C \ B ) ) )
9	6, 7, 8	3bitr4g 303	1 |- ( ( A C_ C /\ B C_ C ) -> ( ( C \ A ) C. B <-> ( C \ B ) C. A ) )

Syntax hints:   -. wn 3   -> wi 4   <-> wb 196   /\ wa 384   \ cdif 3571   C_ wss 3574   C. wpss 3575

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-ne 2795  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-pss 3590

This theorem is referenced by:  isfin2-2  9141  compssiso  9196