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Tuesday, November 16, 2010. If n 5 then n:= n 2. If( n 2=8) or (n-3=6) then n:= 2 * n 1. If( n-3=16) and ([n/6]= 1) then n:= n 3. If( n≠21) and (n-7= 15) then n:= n – 4. N 5 and n=7 (given). N 2 = 9 2 =11 ≠ 8 (false). N – 3 = 9 – 3 = 6 = 6 (true). N/6] = 1 (given). 19/6]= 3 ≠ 1 (false). Hence n = 19. N =19 ≠ 21 (true). N-7 = 15 (given). 19 – 7 =12 ≠ 15 (false). Hence n = 19. QNO14) For primitive statement p,q. Verify that p→[q→(p. Q)] is a tautology. Hence the given compound statement is a tautology.

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Tuesday, November 16, 2010. If n 5 then n:= n 2. If( n 2=8) or (n-3=6) then n:= 2 * n 1. If( n-3=16) and ([n/6]= 1) then n:= n 3. If( n≠21) and (n-7= 15) then n:= n – 4. N 5 and n=7 (given). N 2 = 9 2 =11 ≠ 8 (false). N – 3 = 9 – 3 = 6 = 6 (true). N/6] = 1 (given). 19/6]= 3 ≠ 1 (false). Hence n = 19. N =19 ≠ 21 (true). N-7 = 15 (given). 19 – 7 =12 ≠ 15 (false). Hence n = 19. QNO14) For primitive statement p,q. Verify that p→[q→(p. Q)] is a tautology. Hence the given compound statement is a tautology.

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3 verify that p

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7 q→ p

8 p→ q→ p

9 q→ q→ q

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assignment section,solution,verify that p,is p,q → q→ p,q a tautology,q→ p,p→ q→ p,q→ q→ q,q → q→q,thus p,p x x 0,1 for all,for all,x s x →q x,x s x,x q x,2 for some,x p x,3 for all,4 for all,5 for some,6 for all,r x →s x,b 1 true,2 true,5 true,6 true

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Assignment Section | assignmentsection.blogspot.com Reviews

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Tuesday, November 16, 2010. If n 5 then n:= n 2. If( n 2=8) or (n-3=6) then n:= 2 * n 1. If( n-3=16) and ([n/6]= 1) then n:= n 3. If( n≠21) and (n-7= 15) then n:= n – 4. N 5 and n=7 (given). N 2 = 9 2 =11 ≠ 8 (false). N – 3 = 9 – 3 = 6 = 6 (true). N/6] = 1 (given). 19/6]= 3 ≠ 1 (false). Hence n = 19. N =19 ≠ 21 (true). N-7 = 15 (given). 19 – 7 =12 ≠ 15 (false). Hence n = 19. QNO14) For primitive statement p,q. Verify that p→[q→(p. Q)] is a tautology. Hence the given compound statement is a tautology.

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Tuesday, November 16, 2010. If statement q has the truth value 1, determine all truth value assignments for the primitive statements p,r. For which the truth value of the statement. Q- [( pvr) s]. S- ( r q). Q- ( pvr) s)]- [ s- ( r q)]. P - q) (q - r) (r - p) ]. P- q) (q- r) (r- p) ]. P - q) (q - ) (r - p). P- q) (q- r) (r- p). By comparing column no 7 and 11, we can say that. P - q) (q - r) (r - p) ]. P- q) (q- r) (r- p) ]. 3) let p (x,y) , q (x,y) denote the following open statement,. P (x,y) :. Let B=...

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http://assignmentsection.blogspot.com/2010/11/team-14-ex-2.html

Tuesday, November 16, 2010. If n 5 then n:= n 2. If( n 2=8) or (n-3=6) then n:= 2 * n 1. If( n-3=16) and ([n/6]= 1) then n:= n 3. If( n≠21) and (n-7= 15) then n:= n – 4. N 5 and n=7 (given). N 2 = 9 2 =11 ≠ 8 (false). N – 3 = 9 – 3 = 6 = 6 (true). N/6] = 1 (given). 19/6]= 3 ≠ 1 (false). Hence n = 19. N =19 ≠ 21 (true). N-7 = 15 (given). 19 – 7 =12 ≠ 15 (false). Hence n = 19. QNO14) For primitive statement p,q. Verify that p→[q→(p. Q)] is a tautology. Hence the given compound statement is a tautology.

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Tuesday, November 16, 2010. If n 5 then n:= n 2. If( n 2=8) or (n-3=6) then n:= 2 * n 1. If( n-3=16) and ([n/6]= 1) then n:= n 3. If( n≠21) and (n-7= 15) then n:= n – 4. N 5 and n=7 (given). N 2 = 9 2 =11 ≠ 8 (false). N – 3 = 9 – 3 = 6 = 6 (true). N/6] = 1 (given). 19/6]= 3 ≠ 1 (false). Hence n = 19. N =19 ≠ 21 (true). N-7 = 15 (given). 19 – 7 =12 ≠ 15 (false). Hence n = 19. QNO14) For primitive statement p,q. Verify that p→[q→(p. Q)] is a tautology. Hence the given compound statement is a tautology.

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