Theorem eldifpr 3693

Description: Membership in a set with two elements removed. Similar to eldifsn 3795 and eldiftp 3712. (Contributed by Mario Carneiro, 18-Jul-2017.)

Assertion

Ref	Expression
eldifpr	|- ( A e. ( B \ { C , D } ) <-> ( A e. B /\ A =/= C /\ A =/= D ) )

Proof of Theorem eldifpr

Step	Hyp	Ref	Expression
1		elprg 3686	. . . . 5 |- ( A e. B -> ( A e. { C , D } <-> ( A = C \/ A = D ) ) )
2	1	notbid 671	. . . 4 |- ( A e. B -> ( -. A e. { C , D } <-> -. ( A = C \/ A = D ) ) )
3		neanior 2487	. . . 4 |- ( ( A =/= C /\ A =/= D ) <-> -. ( A = C \/ A = D ) )
4	2, 3	bitr4di 198	. . 3 |- ( A e. B -> ( -. A e. { C , D } <-> ( A =/= C /\ A =/= D ) ) )
5	4	pm5.32i 454	. 2 |- ( ( A e. B /\ -. A e. { C , D } ) <-> ( A e. B /\ ( A =/= C /\ A =/= D ) ) )
6		eldif 3206	. 2 |- ( A e. ( B \ { C , D } ) <-> ( A e. B /\ -. A e. { C , D } ) )
7		3anass 1006	. 2 |- ( ( A e. B /\ A =/= C /\ A =/= D ) <-> ( A e. B /\ ( A =/= C /\ A =/= D ) ) )
8	5, 6, 7	3bitr4i 212	1 |- ( A e. ( B \ { C , D } ) <-> ( A e. B /\ A =/= C /\ A =/= D ) )

Syntax hints:   -. wn 3   /\ wa 104   <-> wb 105   \/ wo 713   /\ w3a 1002   = wceq 1395   e. wcel 2200   =/= wne 2400   \ cdif 3194   {cpr 3667

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-in1 617  ax-in2 618  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-3an 1004  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-ne 2401  df-v 2801  df-dif 3199  df-un 3201  df-sn 3672  df-pr 3673

This theorem is referenced by:  rexdifpr  3694  rplogbval  15619