Some Notes on Exercise 1.2.2

The sentences from exercise 1.2.1, with all possible parentheses dropped.

i. A

Not much to do here.

ii, iii are not wffs

iv. (A→B)

A→B (you can always drop the outside pair)

v is not a wff

vi. (A→(B→C))

A→(B→C): the outside pair can be dropped, but 'A→B→C' is ambiguous.

vii. ((P&Q)→R)

P&Q→R: since '&' is stronger than '→', 'P&Q' must be the wff; if we add parentheses around it, getting '(P&Q)→R', the result is unambiguous.

viii. ((A&B)v(C→(D↔G)))

(A&B)v(C→(D↔G)). We can't drop the parentheses around 'A&B' because 'v' and '&' are equally strong. We can't drop the parentheses around 'C→(D↔G)' because 'v' is stronger than '→' (so we would have to read ((A&B)vC)→...). Likewise, since '→' binds more strongly than '↔', we can't drop the parentheses around 'D↔G'.

ix. ~(A→B)

Nothing to do (if we drop the parentheses, the tilde binds only to 'A').

x. ~(PvQ)v~(QvR)

I said earlier that this isn't a wff because it lacks the outside parentheses; however, it is what you get by dropping all the parentheses you can from a wff. (Sneaky.)

xi, xii are not wffs.

xiii. (~(P&P)v(P↔(Qv~Q)))

First, drop the outside parentheses:
~(P&P)v(P↔(Qv~Q))

The left disjunct '~(P&P)' is a negation; we can't drop the parentheses because tilde binds more strongly than ampersand. The right disjunct is a biconditional, and wedge binds more strongly than double-arrow, so we can't drop the parentheses around it either. However, the right side of the right disjunct is a disjunction, and since wedge binds more strongly than double-arrow we can drop the parentheses around '(Qv~Q)' to get:
~(P&P)v(P↔Qv~Q)

xiv. (~((BvP)&C)↔((Dv~G)→H))

~((BvP)&C)↔Dv~G→H. First drop the outside parentheses:
~((BvP)&C)↔((Dv~G)→H)
Next, since this is a biconditional, and '↔' binds more weakly than any connective, and there's only one '↔' here, we can omit parentheses around its two components, giving
~((BvP)&C)↔(Dv~G)→H
(Note that there weren't any parentheses to drop around its first component.) Next, the right component contains a wedge and an arrow. Since wedge binds more strongly than arrow, we can drop the parentheses around the antecedent of '(Dv~G)→H' :
~((BvP)&C)↔Dv~G→H

xv is not a wff

For 1.2.3, see the answers in Allen/Hand.

To the Syllabus

On to Exercise 1.3

Back to Exercise 1.2.1