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®^

T®^.®^

^®^

T

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)

-(qqr)

-(T q r)            -(^ q r)

^

B does not imply C (fell swoop)

q q r. ® p:®. p ® r

T ®.T®^

T®^

^

A implies C (fell swoop)

q qr.®.p ® r

^      qq^.®.T®^

^®^

T

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

r® qr

q

T^

Page 67 # 2

'p®qr' is equivalent to 'p®q.p®r'

p®qr.«:p®q.p®r

T®qr.«:T®q.T®r        ^®qr.«:^®q.^®r

qr«qr                                   T«T

T                                           T

'p®qr' is not equivalent to 'p®q.q.p®r'

p®qr.«:p®q.q.p®r

T®qr.«:T®q.q.T®r              qr.«:q.q.r

qr«.q qr                                            T«.T q T

Tr«.T q r                                                       T

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

r

T ^

'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

T                                                            T

Page 67 # 4 third pair

pqr q pq q pr q p            pqqqr . pqqq . pqqr . pqq

pq       q        p                              p q q            .            p q

p                                                                p

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.