Advanced Symbolic Logic, PL 330

Homework # 1 Solution

The task is to write a formula using only conjunction and negation plus the statement letters 'p', 'q', and 'r' that will come out true when, but only when, exactly two of the three are true. We might start with something that is equivalent to what we want, namely:

q q

and then apply DeMorgan's law in order to eliminate the alternations. This will take a lot of time. A better strategy is to begin with the eight way alternation of all the possible combinations of our statement letters and then rule some of them out.

pqr q q q q q q q

We want to eliminate the first, fifth, sixth, seventh and eighth options, so we simply negate each of tem and then conjoin the negations:

-(pqr) –() –() –()-()

and there is the answer!

Page 32, problem # 3 First part

p . q q r . ® s : « . pq ® s . pr ® s

and for (p v q)( « (r v p))

p q q : « . r q p

#4 {p ® [(q q r) (p q qs))} « {[q ((q))]®}

#5 (all 3 parts, just because you're special)

ECKpAqrsKCKpqsCCKprs

CKNApqArsKNApqs

ECpKAqrApKqsCAKNqNrKNpANqNsp

Symbolizing “If if if p then q then p then p” as a logical schema in dot notation:

((p → q) → p) → p

p → q. → p: → p