Metamath Proof Explorer


Theorem elsuci

Description: Membership in a successor. This one-way implication does not require that either A or B be sets. (Contributed by NM, 6-Jun-1994)

Ref Expression
Assertion elsuci
|- ( A e. suc B -> ( A e. B \/ A = B ) )

Proof

Step Hyp Ref Expression
1 df-suc
 |-  suc B = ( B u. { B } )
2 1 eleq2i
 |-  ( A e. suc B <-> A e. ( B u. { B } ) )
3 elun
 |-  ( A e. ( B u. { B } ) <-> ( A e. B \/ A e. { B } ) )
4 2 3 bitri
 |-  ( A e. suc B <-> ( A e. B \/ A e. { B } ) )
5 elsni
 |-  ( A e. { B } -> A = B )
6 5 orim2i
 |-  ( ( A e. B \/ A e. { B } ) -> ( A e. B \/ A = B ) )
7 4 6 sylbi
 |-  ( A e. suc B -> ( A e. B \/ A = B ) )