Description: There is exactly one element in a singleton. Exercise 2 of TakeutiZaring p. 15 (generalized). (Contributed by NM, 13-Sep-1995) (Proof shortened by Andrew Salmon, 29-Jun-2011)
Ref | Expression | ||
---|---|---|---|
Assertion | elsng | |- ( A e. V -> ( A e. { B } <-> A = B ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | eqeq1 | |- ( x = A -> ( x = B <-> A = B ) ) |
|
2 | df-sn | |- { B } = { x | x = B } |
|
3 | 1 2 | elab2g | |- ( A e. V -> ( A e. { B } <-> A = B ) ) |