Some Notes on Exercise 1.2.1

The rules of formation that define the language are recursive: roughly speaking, they can be applied to their own results to yield increasingly complicated formulas. This can be continued indefinitely far. However, each non-atomic wff is still constructed in a unique series of steps from other wffs, and each step involves the application of just one rule. It's the last rule used in constructing the wff that defines what kind of wff it is:

To tell whether an expression is a wff, you try to apply rules to it in reverse. Here' s Exercise 1.1, with some comments to show how to do this.

i. A
This is a wff by rule 1. It's also an atomic sentence.
ii. (A
This isn't a wff; the only rules that introduce parentheses are 3, 4, 5, and 6, and they all introduce pairs.
iii. (A)
This isn't a wff either; the only rules introducing parentheses are 3, 4, 5, and 6, and they all require parentheses be put around two wffs connected by '&', 'v', '->', or '<->'.
iv. (A->B)
By rule 5, this is a wff if A and B are wffs; by rule 1, A and B are both wffs. Since rule 5 is the only rule used, it's a conditional.
v. (A->(
Not a wff because '->' must always have a wff on each side (among other things), with parentheses around the whole; '(' is not a wff.
vi. (A->(B->C))
This has parentheses around it, so it's a wff if the result of taking off the parentheses , 'A->(B->C)', is two wffs with a connective between. But 'A' is a wff (rule 1); and '(B->C)' is also a wff (rule 5); so the whole thing is a wff (again by rule 5). Since rule 5 is the last rule used, it's a conditional.
vii. ((P&Q)->R)
As in the last example, this is a wff (by rule 5) if '(P&Q)' and 'R' are wffs. The first is a wff (in fact a conjunction), and the second is a wff (and atomic). So, the entire expression is a wff (and a conditional).
viii. ((A&B) v (C->(D<->G)))
This one is more complicated. First, take off the outside pair of parentheses; this gives:
(A&B)
v
(C->(D<->G))
By rule 3, that will be a wff if '(A&B)' and '(C->(D<->G)' are wffs. So, if it's anything, it's a disjunction. Obviously, '(A&B)' is a wff. '(C->(D<->G)' without its outside parentheses becomes 'C' , '->' , and '(D->G)'. The first is atomic and the last is a conditional, so the whole expression is a wff. Therefore, working back to the beginning, since both '(A&B)' and '(C->(D<->G)' are wffs, the original expression is a wff (and a disjunction).
ix. ~(A->B)
This consists of '(A->B)' , which is a wff (rule 5), preceded by '~', so it's a wff by rule 2 (and therefore a negation).
x. ~(P->Q) v ~(Q&R)
This one can be misleading. Let's start from the left. The only rule for introducing '~' is rule 2, which says that a tilde followed by a wff is a wff. So, if we find a '~', we have to find a wff after it. What follows the first '~'? The entire expression is ' (P->Q) v ~(Q&R)'. That begins and ends with parentheses, so it will be a wff if 'P->Q) v ~(Q&R' is two wffs with a connective between. Unfortunately, it isn't: 'P->Q)' and '(Q&R' aren't wffs. So, whatever the whole expression is, it can't be a negation. However, you may have noticed that '~(P->Q)' by itself is a wff, since it is the wff '(P->Q)' preceded by '~'. in the same way, '~(Q&R)' is a wff. The whole expression, then, consists of two wffs with a 'v' in between. Unfortunately, that's not a wff.
xi. ~(A)
This would be a negation if '(A)' were a wff, but it isn't (see iii above).
xii. (~A)->B
Not a wff because: (1) parentheses need to surround the wffs connected by '->'; (2) '(~A)' is not a wff.
xiii. (~(P&P)&(P<->(Qv~Q)))
This is a wff, and in fact a conjunction. The following table may help show how it's put together (provided this works in your browser!):
(~(P&P)&(P<->(Qv~Q)))
~(P&P)(P<->(Qv~Q))
(P&P) P(Qv~Q)
P P ~ Q
xiv. (~((BvP)&C)<->((DvG)->H))
Biconditional with left side '~((BvP)&C)', right side '((DvG)->H)'. The left side, in turn, is the negation of '((BvP)&C)', and that in turn is a conjunction with conjuncts '(BvP)' and 'C' (and one of those is compound, too). The right side is a conditional with a disjunction for its antecedent and an atomic sentence for its consequent. And you should practice reading that whole description until it makes perfect sense.
xv. (~(Qv~(B))v(E<->(DvX)))
Not a wff because of the expression '(B)', which isn't a wff.