Proposition Logic

Question

Comments

Always true

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It is always true...in order to solve these question, best method to apply like solving digital logic problems....you will get tautology here...as it is the easy way to solve these kind of questions.

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is there any other way to approach this apart from truth table?

Answer 1

$P\leftrightarrow(Q\vee\neg Q)$

$P\leftrightarrow True$

means,the P has to be true i.e. P=1.

$Q\leftrightarrow R$ holds means $Q=R$.

now the expression $(P\wedge Q) \implies ((P\wedge R)\vee S) $ $\equiv$ $(P.Q)\implies((P.R)+S)$

put P=1,  in above expression

$1.Q\implies((1.R)+S)$

$Q\implies R+S$

$Q'+R+S$ $\equiv$$R'+R+S$ $\equiv$$1+S$ $\equiv$ $1$

$\text{So the above expression is true}$

Comments

Q⟹R+SQ⟹R+S

Q′+R+S  how?

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@manisha

$P\implies Q$  $\equiv $   $\neg P \vee Q$ and in

Digital logic we can write it as $P'+Q$

Answer 2

 u can take it as:

(P*Q)-->(PQ)+S (TAKE UNION AS + AND INTERSECTION AS *)

=(PQ)'+(PQ)+S (AS A-->B <==> A'+B)

=P'+Q'+PQ+S 

=P'+PQ+Q'+S (AS A+BC <==> (A+B)(A+C)

=(P'+P)(P'+Q)+Q'+S

=P'+Q+Q'+S

=1+P'+S====>1

IF THIS VALUE COMES OUT TO BE

@1 ITS TAUTOLOGY(TRUE)

@0 ITS CONTRADICTION(FALSE)

@ANY VARIABLE (CONTINGENCY)

and here value is 1 so its a Tautology or True(A)