Theorem pssdifcom1 3636

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 3635	. . . 4 |- ((C \ A) C_ B <-> (C \ B) C_ A)
2	1	a1i 10	. . 3 |- ((A C_ C /\ B C_ C) -> ((C \ A) C_ B <-> (C \ B) C_ A))
3		ssconb 3400	. . . . 5 |- ((B C_ C /\ A C_ C) -> (B C_ (C \ A) <-> A C_ (C \ B)))
4	3	ancoms 439	. . . 4 |- ((A C_ C /\ B C_ C) -> (B C_ (C \ A) <-> A C_ (C \ B)))
5	4	notbid 285	. . 3 |- ((A C_ C /\ B C_ C) -> (-. B C_ (C \ A) <-> -. A C_ (C \ B)))
6	2, 5	anbi12d 691	. 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 3356	. 2 |- ((C \ A) C. B <-> ((C \ A) C_ B /\ -. B C_ (C \ A)))
8		dfpss3 3356	. 2 |- ((C \ B) C. A <-> ((C \ B) C_ A /\ -. A C_ (C \ B)))
9	6, 7, 8	3bitr4g 279	1 |- ((A C_ C /\ B C_ C) -> ((C \ A) C. B <-> (C \ B) C. A))

Syntax hints:  -. wn 3   -> wi 4   <-> wb 176   /\ wa 358   \ cdif 3207   C_ wss 3258   C. wpss 3259

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ne 2519  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-dif 3216  df-ss 3260  df-pss 3262

This theorem is referenced by: (None)