theorem sbneq1d {x: nat} (G: wff x) (a b c: nat x):
$ G -> a = b $ >
$ G -> N[a / x] c = N[b / x] c $;
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | sbneq1 | a = b -> N[a / x] c = N[b / x] c |
|
| 2 | hyp h | G -> a = b |
|
| 3 | 1, 2 | syl | G -> N[a / x] c = N[b / x] c |