GO Classes 2023 | Weekly Quiz 3 | Question: 9

Answer 1

We have
$p\Rightarrow (q\Rightarrow r)\;\Leftrightarrow \sim p\vee (\sim q \wedge r)$
$\quad \quad \quad \quad \quad \quad \Leftrightarrow \sim p\wedge \sim q \wedge r$
$\text{and}$
$(p\wedge q)\Rightarrow r\;\;\; \Leftrightarrow \sim (p\wedge q)\vee r$
$\quad \quad \quad \quad \quad \quad \Leftrightarrow \sim p \wedge \sim q \wedge r$
Hence the expression in large parentheses in the problem is always true. Thus the problem reduces to rewriting $\sim p \wedge \sim q \wedge \sim r$. By DeMorgan's law, this equals $\sim (p\vee q\vee r).$

Comments

Nothing is given about p, q, and r. How can we conclude if S is tautology, contingency, or contradiction?

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For p, q, r, each have only two cases possible either true or false.

Assume (p, q, r) is true, then ~(p v q v r)  is false.

Now in case (p, q, r) is false, then ~(p v q v r)  is true.

So it is contingency

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@ashutoshkpandey but it's not given if p,q, and r are atomic propositions or propositional formulae. Depending on what those are, S can be always true or always false as well.

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In answer this must be,

p → (q → r)    ↔    ~p v ~q v r

Answer 2

Answer C&D

Let Two Cases for P

 

1^st Case P=True

 S :   (((q⇒r))⇔((q)⇒r))∧∼p∧∼q∧∼r   As (True⇒Anything is Anything) and (True AND Anything is anything)

S :    ((q⇒r)⇔(q⇒r))∧∼p∧∼q∧∼r        ((q⇒r)⇔(q⇒r)) is always True As (P⇔P) is always True

S :    T∧∼p∧∼q∧∼r

S :    ∼p∧∼q∧∼r Option D

and p is True so ∼p is False. Then S is a Contradiction

 

2^nd Case P=False

 S :   ((p⇒(q⇒r))⇔((p∧q)⇒r))∧∼p∧∼q∧∼r

False⇒Anything is True and False∧Anything is False

S :   (T⇔(F⇒r))∧∼p∧∼q∧∼r

S :   (T⇔T)∧∼p∧∼q∧∼r

S :   T∧∼p∧∼q∧∼r

S :   ∼p∧∼q∧∼r Option D

And P is False so ∼p is True. Then S :  ∼q∧∼r which is a contingency Option C