Theorem eldiftp 3601

Description: Membership in a set with three elements removed. Similar to eldifsn 3682 and eldifpr 3583. (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 3107	. 2 |- ( A e. ( B \ { C , D , E } ) <-> ( A e. B /\ -. A e. { C , D , E } ) )
2		eltpg 3600	. . . . 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 2412	. . . 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 962   /\ w3a 963   = wceq 1332   e. wcel 2125   =/= wne 2324   \ cdif 3095   {ctp 3558

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-3or 964  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  df-tp 3564

This theorem is referenced by: (None)