1) If statement q has the truth value 1, determine all truth value assignments for the primitive statements p,r and s for which the truth value of the statement
(q -> [ (~p V r) ^ ~s ] ) ^ [ ~s -> ( ~r ^ q ) ] is 1
| P | r | s | ~p | ~pVr | ~s | (~pvr)^~s | q-> [(~pvr)^~s] | ~r | ~r^q | ~s-> (~r^q) | [q->((~pvr)^~s)]-> [~s->(~r^q)] |
| 1 | 1 | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
| 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 1 | 0 |
| 0 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 0 |
| 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
From the above truth table ,
P=0, r=0 and s=0
2) Verify that [ (p<->q) ^ (q<->r) ^ (r<->p) ] ó [ (p->q) ^ (q->r) ^ (r->p) ]
| p | Q | r | P<->q | q<->r | r<->p | (p<->q)^(q<->) ^(r<->p) | p->q | q->r | r->p | (p->q)^(q->r)^(r->p) |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 1 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
| 1 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 1 | 1 | 0 |
| 0 | 1 | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 1 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 |
| 0 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 0 | 0 |
| 0 | 0 | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
By comparing column no 7 and 11, we can say that
[ (p<->q) ^ (q<->r) ^ (r<->p) ] ó [ (p->q) ^ (q->r) ^ (r->p) ]
3) let p (x,y) , q (x,y) denote the following open statement,
p (x,y) : x² ≥ y
q (x,y) : x+2 < y
if the universe for each of x , y consists of all real numbers , determine the truth value for each of the following statements.
è
Given : p (x,y) : x² ≥ y
q (x,y) : x+2 < y
where x , y € R.
A) p (2,4)
è 2² ≥ 4
i.e 4 = 4
.’. P (x,y) IS TRUE.
B) q(1,Π )
è 1+2 < Π
i.e 3<Π
.’. q (1,Π ) IS TRUE . [ ‘.’ Π=3.142]
C) p (-3 ,8 ) ^ q (1 ,3)
è ((-3)² ≥ 8) ^ (1+2 < 3)
(9≥8) ^ (3<3)
TRUE ^ FALSE
.’. FALSE
D) p (1/2 ,1/3) V ~q (-2 ,-3)
è (1/4 ≥ 1/3) V (~ ( -2 + 2 < -3))
FALSE V ~ ( FALSE )
FALSE V TRUE
.’. TRUE.
E) p ( 2,2 )àq ( 1,1 )
è (4 ≥ 2)à (3 < 1)
TRUEàFALSE
.’. FALSE.
F) p (1 ,2)ßà (~ q(1 ,2))
è (1 ≥ 2)ßà(~ ( 3 < 2 ))
FALSE ßà (~( FALSE))
FALSE ßà TRUE
.’. FALSE
4)
a) how many sub sets of {1,2,3,……11} contains at least one even integer.
b) how many sub sets of {1,2,3,……12} contains at least one even integer.
c) generalize the results of part (a) and (b)
è
We know that ,
Number of sub sets of a set X is given by 2ⁿ.
Where ‘n’ is the number of elements in X. i.e n = |X|.
a) let A={1,2,3,…….11}
here |A| = 11,
,’. Total Sub sets of A = 2¹¹
= 2048.
Now considering only odd integers in A,
|A|(odd)=6
.’. Sub sets of odd integers of A = 64
.’. sub sets containing at least one even integer is given by
A(even) = total sub sets of A – sub sets with only odd integers
= 2048 – 64
= 1984.
b) let B={1,2,3,…….12}
here |B| = 12,
,’. Total Sub sets of B = 2¹²
= 4096.
Now considering only odd integers in B,
|B|(odd)=6
.’. Sub sets of odd integers of B = 64
.’. sub sets containing at least one even integer is given by
B(even) = total sub sets of B – sub sets with only odd integers
= 4096 – 64
= 4032.
c) In general let the number of even integers in the given set be n then,
If n is even, then
total number of sub sets contains at least one even integer.=[2n-2n/2]
If n is odd, then
total number of sub sets contains at least one even integer.=[2n-2(n+1)/2]
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