Let p and q be two propositions. Then the implication ∼(p↔q) is :

**A. ** ∼p∨∼q

**B. ** ∼p∧∼q

**C. ** None of these

**D. ** (p∧∼q)∨(∼p∧q)

**Answer : ****Option D**

**Explaination / Solution: **

∼(p↔q)≡(p∧∼q)∨(q∧∼p)

∼(p↔q)≡(p∧∼q)∨(q∧∼p)

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Let p be the proposition : Mathematics is interesting and let q be the proposition that Mathematics is difficult, then the symbol p∧q means

**A. ** Mathematics is interesting and Mathematics is difficult

**B. ** Mathematics is interesting or Mathematics is difficult

**C. ** Mathematics is interesting implies that Mathematics is difficult

**D. ** Mathematics is interesting implies and is implied by Mathematics is difficult

**Answer : ****Option A**

**Explaination / Solution: **

using connective and for ∧

using connective and for ∧

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Let p and q be two prepositions given by p : It is hot, q : He wants water Then , the verbal meaning of p→q is

**A. ** If and only if it is hot, he wants water.

**B. ** it is hot or he wants water.

**C. ** If it is hot , then he wants water.

**D. ** it is hot and he wants water.

**Answer : ****Option C**

**Explaination / Solution: **

→ symbol is replaced by if p then q

→ symbol is replaced by if p then q

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Let p and q be two prepositions given by p : I have the raincoat, q : I can walk in rain. The compound proposition “ If I have the raincoat , then I can walk in the rain “ is represented by

**A. ** p ∨q

**B. ** p↔q

**C. ** p→q

**D. ** p ∨q

**Answer : ****Option C**

**Explaination / Solution: **

If then means an implication statement .hence we rewrite as p→q

If then means an implication statement .hence we rewrite as p→q

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Which of the following is logically equivalent to (p∧q) ?

**A. ** ∼(p∧∼q)

**B. ** ∼(∼pv∼q)

**C. ** p→∼q

**D. ** (∼p∧∼q)

**Answer : ****Option B**

**Explaination / Solution: **

∼(∼(p∧q)) since∼p∨∼q≡∼(p∧q) ∼(∼p)≡p

∼(∼(p∧q)) since∼p∨∼q≡∼(p∧q) ∼(∼p)≡p

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Which of the following is logically equivalent to p↔q ?

**A. ** (p∧q)∧(q→p)

**B. ** (p→q)∧(q→p)

**C. ** (p→q)∨(q→p)

**D. ** (p∧q)∧(q∨p)

**Answer : ****Option B**

**Explaination / Solution: **

By definition p↔q=(p→q)∧(q→p)

By definition p↔q=(p→q)∧(q→p)

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(p∧∼q)∧(∼p∨q) is

**A. ** both a tautology and a contradiction

**B. ** a tautology

**C. ** neither a tautology nor a contradiction

**D. ** a contradiction

**Answer : ****Option D**

**Explaination / Solution: **

Since

F V F = F Since

Hence contracdiction

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Negation of the statement q∨∼(p∧r) is

**A. ** ∼q∧∼(p∧r)

**B. ** ∼q∨(p∧r)

**C. ** ∼q∧(p∧r)

**D. ** ∼q→∼(p∧r)

**Answer : ****Option C**

**Explaination / Solution: **

∼(q∨∼(p∧r)) Since ∼(q∨r)≡∼q∧∼r

∼q∧(p∧r)

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The contrapositive of the statement “ if then I get first class” is

**A. ** If I do not get a first class , then
**B. ** none of these

**C. ** If I do not get a first class , then
**D. ** If I get a first class , then
**Answer : ****Option C**

**Explaination / Solution: **

p:

q:I get first class

the contrapositive of . hence the answer is If I do not get a first class , then

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The inverse of the proposition (p∧∼q)→r is

**A. ** (∼p∨q)→∼r

**B. ** None of these

**C. ** ∼r→∼p∨q

**D. ** r→p∧∼q

**Answer : ****Option A**

**Explaination / Solution: **

Since

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