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

                T .T^




A implies C (fell swoop)

q qr..p r

^      qq^..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

pr.. q qr

Tr.. ^q qr       

       r qr



Page 67 # 2

'pqr' is equivalent to ''

Tqr.:Tq.Tr        ^qr.:^q.^r

qrqr                                   TT

T                                           T


'pqr' is not equivalent to ''

Tqr.:Tq.q.Tr              ^qr.:^q.q.^r

qr.q qr                                            T.T q T

Tr.T q r                                                       T




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

p.q q

T.q q r::Tq.Tr             ^.q q r::^q.^r

q q r.qr                                                   T

                                    T q r.Tr


                                          T ^


'p.q q r' is equivalent to ''

p.q q

T.q q r::Tq.q.Tr           ^.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.