Theorem eldiftp 3679

Description: Membership in a set with three elements removed. Similar to eldifsn 3760 and eldifpr 3660. (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 3175	. 2 |- ( A e. ( B \ { C , D , E } ) <-> ( A e. B /\ -. A e. { C , D , E } ) )
2		eltpg 3678	. . . . 5 |- ( A e. B -> ( A e. { C , D , E } <-> ( A = C \/ A = D \/ A = E ) ) )
3	2	notbid 669	. . . 4 |- ( A e. B -> ( -. A e. { C , D , E } <-> -. ( A = C \/ A = D \/ A = E ) ) )
4		ne3anior 2464	. . . 4 |- ( ( A =/= C /\ A =/= D /\ A =/= E ) <-> -. ( A = C \/ A = D \/ A = E ) )
5	3, 4	bitr4di 198	. . 3 |- ( A e. B -> ( -. A e. { C , D , E } <-> ( A =/= C /\ A =/= D /\ A =/= E ) ) )
6	5	pm5.32i 454	. 2 |- ( ( A e. B /\ -. A e. { C , D , E } ) <-> ( A e. B /\ ( A =/= C /\ A =/= D /\ A =/= E ) ) )
7	1, 6	bitri 184	1 |- ( A e. ( B \ { C , D , E } ) <-> ( A e. B /\ ( A =/= C /\ A =/= D /\ A =/= E ) ) )

Syntax hints:   -. wn 3   /\ wa 104   <-> wb 105   \/ w3o 980   /\ w3a 981   = wceq 1373   e. wcel 2176   =/= wne 2376   \ cdif 3163   {ctp 3635

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 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-3or 982  df-3an 983  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-ne 2377  df-v 2774  df-dif 3168  df-un 3170  df-sn 3639  df-pr 3640  df-tp 3641

This theorem is referenced by: (None)