(4 b is offering some resistance)
I get a "contradictory" solution again.
1. Forall-x (~P(x) && Q(x)) ...... premise
2. box-1 ... The arbitrary x
0 plus ~P(x
0) && Q(x
0) ........ forall e, 1
3. ... ~P(x
0) .......... && e, 2
4. ... box-2 .. P(x
0) .............. assumption.
5. ... .... "bottom" ......... ~e 3,4
6. ... .... Q(x
0) ......... "bottom" e, 5 .... end of box-2.
7. P(x
0) -> Q(x
0) .......... -> i, 4-6
8. Forall-x (P(x) -> Q(x)) .......... forall-x i 2-7
(For some arbitrary x-value we get the implication. Then we must get it for all x-values).