Theorem eldifpr 3610

Description: Membership in a set with two elements removed. Similar to eldifsn 3710 and eldiftp 3629. (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 3603	. . . . 5 |- ( A e. B -> ( A e. { C , D } <-> ( A = C \/ A = D ) ) )
2	1	notbid 662	. . . 4 |- ( A e. B -> ( -. A e. { C , D } <-> -. ( A = C \/ A = D ) ) )
3		neanior 2427	. . . 4 |- ( ( A =/= C /\ A =/= D ) <-> -. ( A = C \/ A = D ) )
4	2, 3	bitr4di 197	. . 3 |- ( A e. B -> ( -. A e. { C , D } <-> ( A =/= C /\ A =/= D ) ) )
5	4	pm5.32i 451	. 2 |- ( ( A e. B /\ -. A e. { C , D } ) <-> ( A e. B /\ ( A =/= C /\ A =/= D ) ) )
6		eldif 3130	. 2 |- ( A e. ( B \ { C , D } ) <-> ( A e. B /\ -. A e. { C , D } ) )
7		3anass 977	. 2 |- ( ( A e. B /\ A =/= C /\ A =/= D ) <-> ( A e. B /\ ( A =/= C /\ A =/= D ) ) )
8	5, 6, 7	3bitr4i 211	1 |- ( A e. ( B \ { C , D } ) <-> ( A e. B /\ A =/= C /\ A =/= D ) )

Syntax hints:   -. wn 3   /\ wa 103   <-> wb 104   \/ wo 703   /\ w3a 973   = wceq 1348   e. wcel 2141   =/= wne 2340   \ cdif 3118   {cpr 3584

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ne 2341  df-v 2732  df-dif 3123  df-un 3125  df-sn 3589  df-pr 3590

This theorem is referenced by:  rexdifpr  3611  rplogbval  13657