Theorem eldifsni 3520

Description: Membership in a set with an element removed. (Contributed by NM, 10-Mar-2015.)

Assertion

Ref	Expression
eldifsni	|- ( A e. ( B \ { C } ) -> A =/= C )

Proof of Theorem eldifsni

Step	Hyp	Ref	Expression
1		eldifsn 3519	. 2 |- ( A e. ( B \ { C } ) <-> ( A e. B /\ A =/= C ) )
2	1	simprbi 269	1 |- ( A e. ( B \ { C } ) -> A =/= C )

Syntax hints:   -> wi 4   e. wcel 1434   =/= wne 2246   \ cdif 2971   {csn 3400

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-bndl 1440  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064

This theorem depends on definitions:  df-bi 115  df-tru 1288  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-nfc 2209  df-ne 2247  df-v 2604  df-dif 2976  df-sn 3406

This theorem is referenced by:  neldifsn  3521  suppssfv  5733  suppssov1  5734  en2other2  6512