Advanced Symbolic Logic, PL 330
Homework set # 4, answer key
Page 59 # 1
Yes, 'pqq.®. -(st) q uv' implies 'q®' as shown by fell swoop
pqq.®. -(st) q uv :®. q®
T®. ^ q ^ :®. TT®^
Page 59 # 2
Symbolize the three statements as A) 'q qr' B) 'q q r. ® p' and C) ' p ® r'.
A does not imply B
q qr.®:q q r. ® p
^ q qr.®:q q r.® T T q qr.®: q q r. ®^
T T ® -(q q r)
-(T q r) -(^ q r)
B does not imply C (fell swoop)
q q r. ® p:®. p ® r
A implies C (fell swoop)
q qr.®.p ® r
B cannot imply A, for A implies C and if B implied A then by (ii) p 47, B would imply C, which we have shown not to be the case.
Similarly, C cannot imply B, for if C implied B, then, since A implies C, A would, by (ii) p 47 imply B, which it does not.
C does not imply A
p®r.®. q qr
T®r.®. ^q qr
Page 67 # 2
'p®qr' is equivalent to 'p®q.p®r'
'p®qr' is not equivalent to 'p®q.q.p®r'
qr«.q qr T«.T q T
Tr«.T q r T
'p®.q q r' is not equivalent to 'p®q.p®r'
p®.q q r:«:p®q.p®r
T®.q q r:«:T®q.T®r ^®.q q r:«:^®q.^®r
q q r.«qr T
T q r.«Tr
'p®.q q r' is equivalent to 'p®q.q.p®r'
p®.q q r:«:p®q.q.p®r
T®.q q r:«:T®q.q.T®r ^®.q q r:«:^®q.q.^®r
qqr.«.qqr T«.T qT
Page 67 # 4 third pair
pqr q pq q pr q p pqqqr . pqqq . pqqr . pqq
pq q p p q q . p q
Yes, 'p' is equivalent to 'p'
Page 67, # 8, Justification of (ix), p. 63
First Part--Justifying that if the alternation of S1 with S2 implies S, then each of S1 and S2 imply S.
Suppose S1 v S2 implies S.
Since S1 clearly implies S1 v S2, then, by rule (ii) of implication (see page 40), S1 implies S through the transitivity of implication. By parity of reasoning, S2 implies S.
Second Part--Justifying that if S1 and S2 each imply S, then the alternation of S1 with S2 implies S.
Suppose S1 and S2 each imply S.
By (vii) then, S is equivalent to each of S1 v S and S2 v S. But then S is equivalent to the alternation of all three, i.e., to S1 v S2 v S. Given this equivalence, however, (vii) yields the result that S1 v S2 implies S. Should the equivalence of S to S1 v S2 v S be in doubt, recall that the third law of interchange, p. 55, preserves equivalence when equivalents are interchanged. Put S2 v S for its equivalent S in S1 v S.
The result is S1 v S2 v S.