∼(∼p)↔p is

**A. ** a tautology

**B. ** a contradiction

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

**D. ** none of these

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

**Explaination / Solution: **

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Which of the following is a preposition?

**A. ** Delhi is on the Jupiter

**B. ** I am an advocate

**C. ** None of these

**D. ** A half open door is half closed

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

**Explaination / Solution: **

It is a false statement other options are open sentences

It is a false statement other options are open sentences

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Let p and q be two prepositions given by p : I take medicine, q : I can sleep then ,the compound statement ∼p→q means

**A. ** If I do not take medicine , then I can sleep.

**B. ** I take medicine iff I can sleep.

**C. ** I take medicine if I can sleep.

**D. ** If I do not take medicine , then I cannot sleep.

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

**Explaination / Solution: **

No Explaination.

No Explaination.

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The logically equivalent propositions of p↔q is

**A. ** (p∧q)→(q∨p)

**B. ** (p→q)∨(q→p)

**C. ** (p→q)∧(q→p)

**D. ** (p∧q)∧(q∨p)

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

**Explaination / Solution: **

p↔q≡(p→q)∧(q→p)

p↔q≡(p→q)∧(q→p)

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Negation of the statement ∼p→(q∨r) is

**A. ** ∼p∧(∼q∧∼r)

**B. ** p∧(q∨r)

**C. ** p∨(q∧r)

**D. ** p→∼(q∨r)

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

**Explaination / Solution: **

rules of negation ∼(p→q)≡p∧∼q Hence ∼p∧(∼q∧∼r)

rules of negation ∼(p→q)≡p∧∼q Hence ∼p∧(∼q∧∼r)

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The compound statement p→(∼p∨q) is false , then the truth values of p and q are respectively

**A. ** T,T

**B. ** T,F

**C. ** F,T

**D. ** F,F

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

**Explaination / Solution: **

since T->F is false.

Hence q=F.

So p=T and q=F

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Which of the following statement is a tautology

**A. ** (∼p∨∼q)→(p∨q)

**B. ** (∼p∨∼q)∨(p∨q)

**C. ** (∼p∨q)∼(p∨∼q)

**D. ** (p∨∼q)∧(p∨q)

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

**Explaination / Solution: **

Since

Hence tautology

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The negation of the proposition “if a quadrilateral is a square, then it is a rhombus “ is

**A. ** if a quadrilateral is not a square , then it is a rhombus

**B. ** if a quadrilateral is a square , then it is not a rhombus

**C. ** a quadrilateral is a square and it is not a rhombus

**D. ** a quadrilateral is not a square and it is a rhombus

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

**Explaination / Solution: **

rules of negation ∼(p→q)≡p∧∼q

rules of negation ∼(p→q)≡p∧∼q

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“If the figure is a rhombus then the diagonals are perpendicular “. The contrapositive of the above statement is

**A. ** If the diagonals are perpendicular, then the figure is a rhombus

**B. ** If the diagonals are not perpendicular, then the figure is not a rhombus

**C. ** If the figure is not a rhombus, then its diagonals are not perpendicular

**D. ** If the diagonals are not perpendicular, then the figure is a rhombus

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

**Explaination / Solution: **

p: the figure is a rhombus q: the diagonals are perpendicular

Contrapositive of

hence If the diagonals are not perpendicular, then the figure is not a rhombus

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Let p and q be two propositions. Then the inverse of the implication p→q is

**A. ** ∼q→p

**B. ** ∼p→q

**C. ** p→∼q

**D. ** ∼p→∼q

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

**Explaination / Solution: **

inverse of p→q≡∼p→∼q

inverse of p→q≡∼p→∼q

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