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 = y -> ( x = B <-> y = B ) ) |
|
| 2 | eqeq1 | |- ( y = A -> ( y = B <-> A = B ) ) |
|
| 3 | df-sn | |- { B } = { x | x = B } |
|
| 4 | 1 2 3 | elab2gw | |- ( A e. V -> ( A e. { B } <-> A = B ) ) |