Theorem eldifpr 3583

Description: Membership in a set with two elements removed. Similar to eldifsn 3682 and eldiftp 3601. (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 3576	. . . . 5 |- ( A e. B -> ( A e. { C , D } <-> ( A = C \/ A = D ) ) )
2	1	notbid 657	. . . 4 |- ( A e. B -> ( -. A e. { C , D } <-> -. ( A = C \/ A = D ) ) )
3		neanior 2411	. . . 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 450	. 2 |- ( ( A e. B /\ -. A e. { C , D } ) <-> ( A e. B /\ ( A =/= C /\ A =/= D ) ) )
6		eldif 3107	. 2 |- ( A e. ( B \ { C , D } ) <-> ( A e. B /\ -. A e. { C , D } ) )
7		3anass 967	. 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 698   /\ w3a 963   = wceq 1332   e. wcel 2125   =/= wne 2324   \ cdif 3095   {cpr 3557

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 604  ax-in2 605  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1481  ax-10 1482  ax-11 1483  ax-i12 1484  ax-bndl 1486  ax-4 1487  ax-17 1503  ax-i9 1507  ax-ial 1511  ax-i5r 1512  ax-ext 2136

This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1335  df-nf 1438  df-sb 1740  df-clab 2141  df-cleq 2147  df-clel 2150  df-nfc 2285  df-ne 2325  df-v 2711  df-dif 3100  df-un 3102  df-sn 3562  df-pr 3563

This theorem is referenced by:  rexdifpr  3584  rplogbval  13201