I am sorry for posting this here, but this is my last resort. I have been fighting with these natural deduction problems for the last two weeks. I take an online college logic course and it makes it difficult to get help with problems with no real instructor. Any help would be greatly appreciated, if not I understand. The problems are:
$\lnot G \lor \lnot B$
$G\implies B$
$\therefore \lnot G$
$B \implies F$
$\therefore U \implies (X \implies F)$
$\lnot G \implies (\lnot H \implies \lnot J)$
$\lnot Z \implies (H \implies J)$
$\lnot (G \vee Z)$
$\therefore H=J$
$D=X$
$D \vee X$
$D \implies (\lnot X \vee \lnot Y)$
$\therefore Y$
$(I \implies E) \implies \lnot V$
$(G \implies E) \implies V$
$\therefore \lnot E$
I have some work done on all of them except the last but have hit roadblocks. One of the more difficult things for me to understand is if you have a premise, and you assign a certain letter to P, can you assign that same letter further on in the premise to Q or R?
Any help is much appreciated Thanks
It might help to walk through these in plain English. Take the first one:
$\lnot G \lor \lnot B$
$G\implies B$
$\therefore \lnot G$
The first line says: Either not G is true or not B is true.
The second line says: If G is true, then B is true as well.
Hence, it cannot be the case that G is true, because if it is, then B is also true. And we have already established in the first line that we must have either not G or not B.
Therefore, the conclusion is: not G.
Edit:
We can also do this more formally. You know $\lnot G \lor \lnot B$ and you want to show $\lnot G$. So assume each of $\lnot G$ and $\lnot B$ and show that they both imply $\lnot G$. Here we go:
Assume $\lnot G$
$\lnot G \implies \lnot G$
$\therefore \lnot G$
Now assume $\lnot B$
Since $G \implies B, \lnot B \implies \lnot G$ (by the contrapositive)
$\therefore \lnot G$
Hence, it either case, $\lnot G$ is true.