1. Using
truth tables show that p → (q Ù r) ≡ (p → q) Ù (p → r).
Answer:-
If p → (q Ù r) «(p → q) Ù (p → r) is a tautology, then p → (q Ù r) ≡ (p → q) Ù (p → r)
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p
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q
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r
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(q Ù r)
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p → (q Ù r)
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(p → q)
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(p → r)
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(p → q) Ù (p → r)
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p → (q Ù r) «(p → q) Ù (p → r)
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T
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T
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T
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T
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T
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T
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T
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T
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T
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T
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T
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F
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F
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F
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T
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F
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F
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T
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T
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F
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T
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F
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F
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F
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T
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F
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T
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T
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F
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F
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F
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F
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F
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F
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F
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T
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F
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T
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T
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T
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T
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T
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T
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T
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T
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F
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T
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F
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F
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T
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T
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T
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T
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T
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F
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F
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T
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F
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T
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T
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T
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T
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T
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F
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F
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F
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F
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T
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T
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T
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T
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T
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p → (q Ù r) «(p → q) Ù (p → r) is a tautology.
Therefore, p → (q Ù r) ≡ (p → q) Ù (p → r)
2. Determine
whether the connective « is
associative.
Answer:-
We have to prove, p «(q «r) = (p «q) «r
If p «(q «r) « (p «q) «r is a tautology, then p «(q «r) = (p «q) «r
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p
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q
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r
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(q «r)
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p «(q «r)
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(p «q)
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(p «q) «r
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p «(q «r) « (p «q) «r
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T
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T
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T
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T
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T
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T
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T
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T
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T
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T
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F
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F
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F
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T
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F
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T
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T
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F
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T
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F
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F
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F
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F
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T
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T
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F
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F
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T
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T
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F
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T
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T
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F
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T
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T
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T
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F
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F
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F
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T
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F
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T
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F
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F
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T
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F
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T
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T
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F
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F
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T
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F
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T
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T
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T
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T
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F
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F
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F
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T
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F
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T
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F
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T
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p «(q «r) « (p «q) «r is a tautology.
Therefore, p «(q «r) = (p «q) «r
The connective « is associative.
3. Show
that the following propositions are tautologies:
i) [¬q Ù (p → q)] → ¬p
ii) [(p → q) Ù (q → r)] → (p → r)
Answer:-
i)
[¬q Ù (p → q)] → ¬p
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p
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q
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¬p
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¬q
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(p → q)
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¬q Ù (p → q)
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[¬q Ù (p → q)] → ¬p
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T
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T
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F
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F
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T
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F
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T
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T
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F
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F
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T
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F
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F
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T
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F
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T
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T
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F
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T
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F
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T
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F
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F
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T
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T
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T
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T
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T
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Therefore, [¬q Ù (p → q)] → ¬p is a tautology.
ii) [(p
→ q) Ù (q → r)] → (p → r)
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p
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q
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r
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(p → q)
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(q → r)
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(p → q) Ù (q → r)
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(p → r)
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[(p → q) Ù (q → r)] → (p → r)
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T
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T
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T
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T
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T
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T
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T
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T
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T
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T
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F
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T
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F
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F
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F
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T
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T
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F
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T
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F
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T
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F
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T
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T
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T
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F
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F
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F
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T
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F
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F
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T
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F
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T
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T
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T
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T
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T
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T
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T
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F
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T
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F
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T
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F
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F
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T
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T
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F
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F
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T
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T
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T
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T
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T
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T
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F
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F
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F
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T
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T
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T
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T
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T
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Therefore, [(p → q) Ù (q → r)] → (p → r) is a tautology.
4. Let A = {Aldgate, Monument, Paddington}
B = {Leicester Square, Victoria, Green
Park}
and let circle(X) be X is
a station on the circle line,
bakerloo(X) be X is a station on the bakerloo line.
Which of the following are true
and which false? Symbolise the negation of each of the following. Your
symbolisation must begin with a quantifier
(i)
"X : A. circle(X)
(ii)
"X : B. bakerloo(X)
(iii)
$X : A. bakerloo(X)
(iv)
$X : B. circle(X)
(v)
$X : B. (circle(X) Ù bakerloo(X))
(vi)
$X : A . Øbakerloo(X)
Answer:-
I.
"X : A. circle(X)
$X : A. Ø circle(X)
II.
"X : B. bakerloo(X)
$X : B. Ø bakerloo(X)
III.
$X : A. bakerloo(X)
"X : A. Ø bakerloo(X)
IV.
$X : B. circle(X)
"X : B. Ø circle(X)
V.
$X : B. (circle(X) Ù bakerloo(X))
"X : B. Ø (circle(X) Ù bakerloo(X))
"X : B. Ø circle(X) V Ø bakerloo(X)
VI.
$X : A . Øbakerloo(X)
"X : A. bakerloo(X)
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