Theorem eldifpr 3621

Description: Membership in a set with two elements removed. Similar to eldifsn 3721 and eldiftp 3640. (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 3614	. . . . 5 |- ( A e. B -> ( A e. { C , D } <-> ( A = C \/ A = D ) ) )
2	1	notbid 667	. . . 4 |- ( A e. B -> ( -. A e. { C , D } <-> -. ( A = C \/ A = D ) ) )
3		neanior 2434	. . . 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 3140	. 2 |- ( A e. ( B \ { C , D } ) <-> ( A e. B /\ -. A e. { C , D } ) )
7		3anass 982	. 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 708   /\ w3a 978   = wceq 1353   e. wcel 2148   =/= wne 2347   \ cdif 3128   {cpr 3595

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 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ne 2348  df-v 2741  df-dif 3133  df-un 3135  df-sn 3600  df-pr 3601

This theorem is referenced by:  rexdifpr  3622  rplogbval  14448