Theorem eldiftp 3622

Description: Membership in a set with three elements removed. Similar to eldifsn 3703 and eldifpr 3603. (Contributed by David A. Wheeler, 22-Jul-2017.)

Assertion

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

Proof of Theorem eldiftp

Step	Hyp	Ref	Expression
1		eldif 3125	. 2 |- ( A e. ( B \ { C , D , E } ) <-> ( A e. B /\ -. A e. { C , D , E } ) )
2		eltpg 3621	. . . . 5 |- ( A e. B -> ( A e. { C , D , E } <-> ( A = C \/ A = D \/ A = E ) ) )
3	2	notbid 657	. . . 4 |- ( A e. B -> ( -. A e. { C , D , E } <-> -. ( A = C \/ A = D \/ A = E ) ) )
4		ne3anior 2424	. . . 4 |- ( ( A =/= C /\ A =/= D /\ A =/= E ) <-> -. ( A = C \/ A = D \/ A = E ) )
5	3, 4	bitr4di 197	. . 3 |- ( A e. B -> ( -. A e. { C , D , E } <-> ( A =/= C /\ A =/= D /\ A =/= E ) ) )
6	5	pm5.32i 450	. 2 |- ( ( A e. B /\ -. A e. { C , D , E } ) <-> ( A e. B /\ ( A =/= C /\ A =/= D /\ A =/= E ) ) )
7	1, 6	bitri 183	1 |- ( A e. ( B \ { C , D , E } ) <-> ( A e. B /\ ( A =/= C /\ A =/= D /\ A =/= E ) ) )

Syntax hints:   -. wn 3   /\ wa 103   <-> wb 104   \/ w3o 967   /\ w3a 968   = wceq 1343   e. wcel 2136   =/= wne 2336   \ cdif 3113   {ctp 3578

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 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-3or 969  df-3an 970  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-ne 2337  df-v 2728  df-dif 3118  df-un 3120  df-sn 3582  df-pr 3583  df-tp 3584

This theorem is referenced by: (None)