Closed Quelklef closed 5 years ago
The following proof is incorrectly marked as invalid:
Ex P(x) | [a] P(a) | Ek P(k)
a workaround is
Ex P(x) | [a] P(a) | P(a) | Ek P(k)
... however, this is a bug and should be fixed.
Seems to be fixed in the latest version
The following proof is incorrectly marked as invalid:
a workaround is
... however, this is a bug and should be fixed.