Discrete Mathematics | Predicate Logic

Question

Which of the following conclusion can be drawn from the following premises

$\left ( 1 \right )\sim P\vee Q\rightarrow R$

$\left ( 2 \right )S\vee \sim Q$

$\left ( 3 \right )\sim T$

$\left ( 4 \right )P\rightarrow T$

$\left ( 5 \right )\sim P\wedge R\rightarrow \sim S$

 

$C1: P$

$C2: \sim Q$

$C3: Q\wedge R$

 

• A. C1 only
• B. C2 only
• C. C2 & C3 only
• D. C1 & C2 only

Answer 1

option B is correct one

CP=contra positive

MP=Modus ponens

Answer 2

Option (B) is correct  

 

 

Comments

So, here we have to check if the conclusion must be true for all premisies to become True or not, right?

like, ~Q must be True for all the premises to become True, thus C2 correct,

but we can make all the premises True even if Q^R is False, Thus C3 in incorrect.

Tell me if my intusion is correct or not?

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YES ,you are correct .It is just like implication statement :

$P_1 \cap P_2 \cap ...\cap P_n \implies conclusion$

where P1 represent premises 1 .If you able to make our conclusion false and premises true then our conclusion is false.(P->Q is false if P=true & Q=false)

Answer 3

Option (b) is correct .

Step To Check whether the conclusion can be drawn from the premise are 

1. Make the Conclusion False
2. Now Try to Make all the premise True 

If We are able to do the above in that case the conclusion is not possible else its possible .

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C1 : P 

Lets Consider the conclusion False So $P =  False$

 

Premise 1 :- $P' \vee Q -> R$

This will Become $True \vee Q → R$ Which is just $True → R$ , Now to make this premise $True$ We have to consider R as $True$

Premise 2 :- $S \vee Q’$

Now To Make This Statement True Lets Assume $S = True$ and $Q’ = True$

Premise 3 :-  $T’$

Now to Make this Statement $True$ Consider $T’ = True $

Premise 4:-  $P → T$

We Have Already Considered $P = False$ in the conclusion and $False → Anything = True $ 

Premise 5 :- $P' \wedge Q → S'$

When we substitute the Value this will become $True \wedge False → False$ and $False – anything = True$

 

Since we are able to make all the premise True and the conclusion False simultaneously This conclusion is false .

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C3 :- $Q \wedge R$

Now to Make this conclusion False $Q = False$ and $R = False$

 

Premise 1 :- $P’ \vee Q → R$

This will become $P’ \vee False → R$ which is nothing but $P’ → R$ now to make this true lets assume $P’ = True$ and $R = True$ . 

Premise 2 :-  $S \vee Q’$

When we substitute the value of $Q’$this premise will become $S \vee True $  which is just $S$ , and now to make this premise True lets assume $S = True$

Premise 3 :-  T’

To Make this  $True$  $T’ = True.$

Premise 4 :- P → T

We Have Previously Considered $P’ = True$  so $P = False$  and  $False → Anything = True .$

Premise 5 :- $P’ \wedge Q → S’$

$P’ = True$ and $Q = False$ , so $P’ \wedge Q$ will become $False$ and $False -> Anything = True$ 

Since we were able to make all the premise True and the conclusion False Simultaneously This conclusion if False .

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Now we can Easily Eliminate Option 1,3,4 , so The Correct Answer is Option :- B