Theorem eqoreldif 4225

Description: An element of a set is either equal to another element of the set or a member of the difference of the set and the singleton containing the other element. (Contributed by AV, 25-Aug-2020.) (Proof shortened by JJ, 23-Jul-2021.)

Assertion

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

Proof of Theorem eqoreldif

Step	Hyp	Ref	Expression
1		simpl 473	. . . . 5 |- ( ( A e. C /\ -. A = B ) -> A e. C )
2		elsni 4194	. . . . . . 7 |- ( A e. { B } -> A = B )
3	2	con3i 150	. . . . . 6 |- ( -. A = B -> -. A e. { B } )
4	3	adantl 482	. . . . 5 |- ( ( A e. C /\ -. A = B ) -> -. A e. { B } )
5	1, 4	eldifd 3585	. . . 4 |- ( ( A e. C /\ -. A = B ) -> A e. ( C \ { B } ) )
6	5	ex 450	. . 3 |- ( A e. C -> ( -. A = B -> A e. ( C \ { B } ) ) )
7	6	orrd 393	. 2 |- ( A e. C -> ( A = B \/ A e. ( C \ { B } ) ) )
8		eleq1a 2696	. . 3 |- ( B e. C -> ( A = B -> A e. C ) )
9		eldifi 3732	. . . 4 |- ( A e. ( C \ { B } ) -> A e. C )
10	9	a1i 11	. . 3 |- ( B e. C -> ( A e. ( C \ { B } ) -> A e. C ) )
11	8, 10	jaod 395	. 2 |- ( B e. C -> ( ( A = B \/ A e. ( C \ { B } ) ) -> A e. C ) )
12	7, 11	impbid2 216	1 |- ( B e. C -> ( A e. C <-> ( A = B \/ A e. ( C \ { B } ) ) ) )

Syntax hints:   -. wn 3   -> wi 4   <-> wb 196   \/ wo 383   /\ wa 384   = wceq 1483   e. wcel 1990   \ cdif 3571   {csn 4177

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-dif 3577  df-sn 4178

This theorem is referenced by:  lcmfunsnlem2  15353