Theorem eldifpr 3649

Description: Membership in a set with two elements removed. Similar to eldifsn 3749 and eldiftp 3668. (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 3642	. . . . 5 |- ( A e. B -> ( A e. { C , D } <-> ( A = C \/ A = D ) ) )
2	1	notbid 668	. . . 4 |- ( A e. B -> ( -. A e. { C , D } <-> -. ( A = C \/ A = D ) ) )
3		neanior 2454	. . . 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 3166	. 2 |- ( A e. ( B \ { C , D } ) <-> ( A e. B /\ -. A e. { C , D } ) )
7		3anass 984	. 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 709   /\ w3a 980   = wceq 1364   e. wcel 2167   =/= wne 2367   \ cdif 3154   {cpr 3623

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 615  ax-in2 616  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-ext 2178

This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-nf 1475  df-sb 1777  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-ne 2368  df-v 2765  df-dif 3159  df-un 3161  df-sn 3628  df-pr 3629

This theorem is referenced by:  rexdifpr  3650  rplogbval  15181