(A)
p | q | ∼p | ∼q | p∨q | p∧q | ∼(p∨q) | ∼p∧q | ∼p∨∼q |
T | T | F | F | T | T | F | F | F |
T | F | F | T | T | F | F | F | T |
F | T | T | F | T | F | F | T | T |
F | F | T | T | F | F | T | F | T |
(B)
p | q | ∼p | ∼q | p∧∼q | p→q | ∼(p→q) | ∼p→∼q |
T | T | F | F | F | T | F | T |
T | F | F | T | T | F | T | T |
F | T | T | F | F | T | F | F |
F | F | T | T | F | T | F | T |
(C)
p | q | p→q | p∧(p→q) | p∧(p→q)→q |
T | T | T | T | T |
T | F | F | F | T |
F | T | T | F | T |
F | F | T | F | T |
2. Prove all the algebraic properties of statements with the help of truth tables.
Solution: Click Here
3. If a, b and c are any three statements, then prove that-
(a) (a∧b)→(a∨b)
(b) [(a→b)∧(b→c)]→(a→c)
are two statements both of which are formula (or tentologies).
Solution:
(a)
a | b | a∧b | a∨b | (a∧b)→(a∨b) |
T | T | T | T | T |
T | F | F | T | T |
F | T | F | T | T |
F | F | F | F | T |
(b)
a | b | c | a→b | b→c | a→c | (a→b)∧(b→c) | [(a→b)∧(b→c)]→(a→c) |
T | T | T | T | T | T | T | T |
T | T | F | T | F | F | F | T |
T | F | T | F | T | T | F | T |
T | F | F | F | T | F | F | T |
F | T | T | T | T | T | T | T |
F | T | F | T | F | T | F | T |
F | F | T | T | T | T | T | T |
F | F | F | T | T | T | T | T |
4. Prove that
(a) ∼(∼p∧∼q)≡p∨q
Solution:
p | q | ∼p | ∼q | p∨q | ∼p∧∼q | ∼(∼p∧∼q) |
T | T | F | F | T | F | T |
T | F | F | T | T | F | T |
F | T | T | F | T | F | T |
F | F | T | T | F | T | F |
(b) ∼(∼p→∼q)≡∼p∨q
Solution:
p | q | ∼p | ∼q | ∼p→∼q | ∼(∼p→∼q) | ∼p∧q |
T | T | F | F | T | F | F |
T | F | F | T | T | F | F |
F | T | T | F | F | T | T |
F | F | T | T | T | F | F |
5. Test if correct or incorrect-
(a) p→q≡∼p→∼q
Solution:
p | q | ∼p | ∼q | p→q | ∼p→∼q |
T | T | F | F | T | T |
T | F | F | T | F | T |
F | T | T | F | T | F |
F | F | T | T | T | T |
(b) ∼(p→q)≡p→∼q
Solution:
p | q | ∼p | ∼q | p→q | ∼(p→q) | p→∼q |
T | T | F | F | T | F | F |
T | F | F | T | F | T | T |
F | T | T | F | T | F | T |
F | F | T | T | T | F | T |
(c) ∼(p→q)≡(p→∼q)∨(q∧∼p)
Solution:
p | q | ∼p | ∼q | p→q | ∼(p→q) | p→∼q | q∧∼p | (p→∼q)∨(q∧∼p) |
T | T | F | F | T | F | F | F | F |
T | F | F | T | F | T | T | F | T |
F | T | T | F | T | F | T | T | T |
F | F | T | T | T | F | T | F | T |
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