theorem sbeq1d {x: nat} (G: wff x) (a b: nat x) (c: wff x):
$ G -> a = b $ >
$ G -> ([a / x] c <-> [b / x] c) $;
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | sbeq1 | a = b -> ([a / x] c <-> [b / x] c) |
|
| 2 | hyp h | G -> a = b |
|
| 3 | 1, 2 | syl | G -> ([a / x] c <-> [b / x] c) |