TEAM 3

1. let p,q be primitive statements for which the implication p->q is false. Determine the truth value for each of the following.

a)p^q

b)~pvq

c)q->p

d)~q->~p

solution:  Given p->q=false

Therefore p=1,q=0;

a)p^q=1^0=0

b)~pvq=0v0=0

c)q->p=0->1=1

d)~q->~p=1->0=0

2.Use the substitution rule to verify that each of the following is a tautology.(Here p,q and r are primitive statements).

a)[p v (q ^r) v ~[p v(q ^r)]

solution: From inverse law we have s v ~s óTo

By replacing each occurrence of s by p v(q ^r) it follows from  the first substitution rule that

[p v(q ^ r) v ~[p v(q ^ r)]óTo

b)[(p v q)->r]ó[~r -> ~(p v q)]

solution: We know s->t ó ~t->~

By replacing each occurrence of s by (p v q) and each occurrence of t by r it follows from first law of substitution that

[(p v q)->r] ó [~r ->~(p v q)]

3. Verify that each of the following is a logical implication by showing that it is impossible for the conclusion to have the truth value 0 while the hypothesis has the truth value 1.

a) (p ^ q)->p: If p has the truth value 0 then p^q is also 0.

b)p->(p v q): If the truth value of p v q is 0 then the truth value of p and that of q will be 0 .

c) [(p v q)   ^ ~p] ->q: If q has truth value of 0, then the truth value of [(pvq) ^ ~p] is 0 regardless of the truth value of p.

d) [(p -> q) ^ (r ->s) ^ (p v r)] -> (q v s): Statement q v s has the truth value 0 only when each of q, s has truth value 0. Then (p -> q) has truth value 1 when p has truth value 0; (r -> s) has truth value 1 when r has truth value 0. But then (p v r) must have truth value 0 not 1.

e) [(p -> q) ^ (r -> s) ^ (~q v ~s)] -> (~p v ~r): Statement (~p v ~r) has the truth value 0 only when each of p, s has truth value 1. Then (p -> q) has truth value 1 when q has truth value 1; (r -> s) has truth value 1 when s has truth value 1. But (~q v ~s) must have truth value 0 not 1.

4. For A= {1, 2, {2}} which of the eight statements given below are true.

a) 1€A

b) {1} €A

c) {2} €A

d) {{1}} C A

e) {2} € A

f) {2} C A

g) {{2}} C A

h) {{2}} C A

solution : a, c, e, f, g, h