Mathematical Reasoning - Online Test

Answer : Option B
Explaination / Solution:

p→(∼p∨q)=F only when p=T and (∼p∨q)=F     since T->F is false.

∼pvq=F only when both∼p and q=F

Hence q=F.

So p=T and q=F

Answer : Option D
Explaination / Solution:

p→q An implication statement,∼q→∼p a contrapositive statement. Implication and contrapositve have same meaning.hence they have identical values. p↔p≡T. hence the above statement will be true always.

Answer : Option A
Explaination / Solution:

The above statement is a fact and the other options are mathematical statements which are proposition .

Answer : Option D
Explaination / Solution:

[(p∧∼q)∧(∼p)]∨[(p∧∼q)∧q)]     Since p∧(q∨r)≡(p∧q)∨(p∧r)

F V F = F     Since       p∧∼p=F

Hence contracdiction

 

Answer : Option D
Explaination / Solution:

p↔∼p   definition of p↔q≡(p→q)∧(q→p)

Hence F

Answer : Option B
Explaination / Solution:

(∼p∨∼q∨p∨q)   Since p∨(q∨r)≡(p∨q)∨r

∼p∨p≡T

Hence tautology

Answer : Option D
Explaination / Solution:

∼(p→(q∧r))          

=p∧∼(q∧r)     since ∼(p→q)≡p∧∼q

=p∧(∼q∨∼r)

Answer : Option A
Explaination / Solution:

(p∧r)∧(∼r∧∼q) since ∼(p→q)≡p∧∼q

Answer : Option C
Explaination / Solution:

Answer : Option A
Explaination / Solution:

∼q→∼p≡q∨∼p        Since p→q≡∼p∨q

∼p∨q≡p→q