TEAM 9
Q.2.1.9 ) which of the compound statements are tautology.
ans )
a] ~(pv~q) -> ~p
p q ~q pv~q ~(pv~q) ~P ~(PV~q)-> ~p
0 0 1 1 0 1 1
0 1 0 0 1 1 1
1 0 1 1 0 0 1
1 1 0 1 0 0 1
above compound statement is tautology.
b] p -> (q -> r)
p q r q -> r p -> (q -> r)
0 0 0 1 1
0 0 1 1 1
0 1 0` 0 1
0 1 1 1 1
1 0 0 1 1
1 0 1 1 1
1 1 0 0 0
1 1 1 1 1
above compound statement is not a tautology.
c] (p -> q) -> r
p q r p -> q (p -> q) -> r
0 0 0 1 0
0 0 1 1 1
0 1 0 1 0
0 1 1 1 1
1 0 0 0 1
1 0 1 0 1
1 1 0 1 0
1 1 1 1 1
it is also not a tautology
d] (p -> q) -> (q -> p)
p q (p -> q) q -> p (p -> q) -> (q -> p)
0 0 1 1 1
0 1 1 0 0
1 0 0 1 1
1 1 1 1 1
it is also not a tautology
e] [ p ^ (p -> q)] -> q
p q p -> q p ^ (p -> q) q [ p ^ (p -> q) ] -> q
0 0 1 0 0 1
0 1 1 0 1 1
1 0 0 0 0 1
1 1 1 1 1 1
the above compound statement is tautology
f] (p ^ q) -> p
p q p ^ q (p ^ q) -> p
0 0 0 1
0 1 0 1
1 0 0 1
1 1 1 1
g] q <-> (~p v ~q)
p q ~ p ~q ~p v ~q q <-> (~p v ~q)
0 0 1 1 1 0
0 1 1 0 1 0
1 0 0 1 1 1
1 1 0 0 0 0
the above compound statement is not a tautology
h] [ (p -> q) ^ (q -> r) ] -> (p -> r)
p q r p -> q q -> r [ (p -> q) ^ (q -> r) ] (p ->r) y
0 0 0 1 1 1 1 1
0 0 1 1 1 1 1 1
0 1 0 1 0 0 1 1
0 1 1 1 1 1 1 1
1 0 0 0 1 0 0 1
1 0 1 o 1 0 1 1
1 1 0 1 0 0 0 1
1 1 1 1 1 1 1 1
hence above statement is tautology.
Q. 2.2.9 ) write the converse,inverse and contrapositive of each of the following implications. for each implication,determine its truth value as well as the truth values of its
corresponding converse,inverse and contrapositive.
ans )
a ) if today is labour day, then tomorrow is tuesday.
the statement is :
p: Today is labour day.
q: Tomorrow is Tuesday.
1.) p -> q
2.) ~q ->p is contrapositive of p ->q
3.) q -> p is converse of p->q
4.) ~p->~q is inverse of p->q
p q ~q ~p ~q->~p q->p ~p->~q p->q
0 0 1 1 1 1 1 1
0 1 0 1 1 0 0 1
1 0 1 0 0 1 1 0
1 1 0 0 1 1 1 1
1.) If today is labour day then tomorrow is tuesday.
2.) If tomorrow is not tuesday then today is not labour day. -> contrapositive -> true
3.) If tomorrow is tuesday then today is labour day -> converse. -> true
4.) If today is not labour day then tomorrow is not tuesday. -> inverse -> true
b ) if -1<3 and 3+7=10 then sin (3#/2) = -1 where # is pie
then statement is
p: -1<3 and 3+7=10
q: sin (3#/2) = -1
then we obtain
p -> q is true
~q -> ~p is true
q -> p is true
~p -> ~q is true
Q.2.4.2 ) let p(x),q(x) be defined as in exercise 1. let r(x) be the open statement " x>0 " once again the universe comprises all integers
determine the truth values of the folllowing statements.
ans. )
a) 1.) p(3) v [ q(3) v ~r(3) ] is true where p(3) is TRUE and q(3) is false and r(3) is true then the negation is false so we can say p(3) v [ q(3) v ~r(3) ] is true.
2. ) p(2) -> [ q(2) -> r(2) ] is true when p(2) is true and q(2) is true and r(2) is true so we can say that is is TRUE.
3. ) [ p(2) ^ q(2) ] -> r(2) is true when p(2) is true and q(2) is true and r(2) is true so we can say that [ p(2) ^ q(2) ] -> r(2) is true.
4. ) p(0) -> [ ~q(-1) <-> r(1) ] is true when p(0) is true and q(-1) is false then ~q(-1) is true and r(1) is true so we can say that it is true.
b. )
[ p(x) ^ q(x) ] ^ r(x) is true for x = 2.
Q. 3.1.9 ) which of the following sets are nonempty ?
ans. )
a. ) A = { x | x E N,2x +7 = 3 }
since |A|= 6
so it is a nonempty set.
b.) B = { x E Z | 3x+5=9 }
since |B| = 7
so it is a non empty set.
c. ) { x | x E Q, pow(x,2) + 4 = 6 }
if B has pow (2,n) subsets of odd cardinality then |B|= n+1.
d. ) { x E R | pow(x,2) + 4 = 6 }
it is not a nonempty set.
e. ) { x E R | pow(x,2) + 3x + 3 = 0 }
it is also nota nonempty set.
f. ) { x | x E C, pow(x,2) + 3x + 3 = 0 }
it is non a nonempty set.
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