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100 votes
100 votes

Consider the following expressions:

  1. $false$
  2. $Q$
  3. $true$
  4. $P\vee Q$
  5. $\neg Q\vee P$

The number of expressions given above that are logically implied by $P \wedge (P \Rightarrow Q)$ is ___________.

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11 Answers

6 votes
6 votes
GIVEN

P->Q

P

________

1.implies Q

2 NOW Q implies P OR Q

3.P IS TRUE so it implies P OR ~Q

4 THIS IS DIFFERENT STILL EASY....P->Q is true ..Q is true and TRUE IS TRUE (trivially)..to understand ..make table of p->T

Answer => 4
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1 votes

For two propositions A and B , if we can take A as true and derive B as true somehow then we can say A logically implies B $( A \implies B)$

Taking given premise A as true

Now check each option whether it is implied by given premise or not

options B, C, D, E are implied by given premise $P \land \left( P \to Q\right)$ and option A doesn’t imply it

$\therefore$ the count is 4

1 votes
1 votes

First Let us simplify :

P˄(P->Q)

=P˄(~P˅Q)

=(P˄~P)˅(P˄Q)

=P˄Q

 

To check whether an expression A logically implies B , we make A → B a tautology , because  logic implication means my conclusion follows from my arguments, so arguments →  conclusion is a tautology 

 

Now , let us check whether (P˄Q) → option(x) is Tautology or not .

 

To check whether an implication is Tautology or not we will use Two Approaches :

 

Approach 1 : Make Premise or LHS of implication True and then try to make Conclusion or RHS of implication False .

“ If we are able to do so then it is not Tautology “ 

“ I we are not able to do so , then it is Tautology “ .

 

Approach 2 : 

Make Conclusion or RHS of implication False and then try to make Premise or LHS of implication True .

“ If we are able to do so then it is not Tautology “ 

“ I we are not able to do so , then it is Tautology “ .

 

Option (i) : P˄Q → False 

Using Approach 2 , 

 RHS = False , Now let us try to make LHS = True 

 P˄Q = True , Taking P = T and Q = T  we are able to to make P˄Q = True .

Since , LHS =True and RHS = False simultaneously , it is not Tautology. 

 

Option (ii) : P˄Q → Q 

Using Approach 2 ,

RHS = Q = False . Now Lets us try to make LHS = True .

LHS = P˄Q . Since Q is False we will not be able to make LHS True. 

Hence , Tautology.

 

Option (iii) : P˄Q → TRUE 

We know that , Anything → True  = True .

So this is Tautology .

 

Option (iv) : P˄Q → P˅Q 

Using Approach 2 ,

RHS =  P˅Q  = False . This will only happen when P = False and Q = False . Now let us make LHS = True .

LHS = P˄Q  . Since P = F and Q =F , we cannot make LHS True .

Hence , Tautology.

 

Option (v) : P˄Q→ ~Q˅P

Using Approach 1 , LHS = P˄Q= True . This will only happen when P =T and Q =T . Now let us make RHS  = False .

RHS = ~Q˅P . Since P = True we cannot make RHS false.

Hence , Tautology . 

 

So , 4 is the answer .

 

 

1 votes
1 votes

Simplify given expression:

$[P \land (P \rightarrow Q)]$
$=P(P’+Q)$
$=PQ$

Logically imply each option:

  1. $PQ \rightarrow false$
    $=(PQ)’ + 0$
    $=P’ + Q’$
    which is NOT valid
     
  2. $PQ \rightarrow Q$
    $=(PQ)’ + Q$
    $=P’ + Q’ + Q$
    $=1$
    hence, valid
     
  3. $PQ \rightarrow true$
    $=(PQ)’ + 1$
    $=1$
    hence, valid
     
  4. $PQ \rightarrow (P+Q)$
    $=(PQ)’ + (P+Q)$
    $=P’ + Q’ + P+Q$
    $=1$
    hence, valid
     
  5. $PQ \rightarrow (Q’ + P)$
    $=(PQ)’ + (Q’ + P)$
    $=P’ + Q’ + Q’ + P$
    $=1$
    hence, valid

No. of expressions logically implied $=4$

Answer:

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