I have been trying to solve this question for couple hours and I still don't get it. The problem I'm trying to solve is 6.14 in the link below.
This is what I attempted so far:
I'm sure this is wrong, but I don't know the right answer.
6.14 is not valid.
The conclusion can be FALSE and the third premise can still be TRUE : it is enough that SameRow(d,f) is FALSE.
BUT if FrontOf(b,f) allows you to derive ¬SameRow(b,f), in this case the argument is valid.
With it and the second premise you are forced to have SameRow(d,f) true and thus, assuming Cube(f) you have the desired contradiction, concluding with ¬Cube(f).