Propositional logic

Question

which one is not tautology?

1.((p v q)^(r v s)) -> ((p^r) v q vs)

2. ((p v q)^(r v s)) -> (q v s)

3.((p v q)^(r v s)) -> (r v q v s)

4.((p v q)^(r v s)) -> (p v q v s)

5.((p v q)^(r v s)) -> (p v q)

please tell how to approach this types of qs,this type of qs came many times in gate.I tried it by making table,but it is not doable..so,help me.

Answer 1

You can do that even without picking pen with the help of boolean algebra 

Note: $p\rightarrow q = \bar{p} \;or\;q = \bar{p}+q$

Comments

rememeber one more thing . p<-> q is basically exnor gate . it can be evaluated as pq+p'q'.

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Praveen Saini sir

Can you please tell the formula which u used to convert $\bar{p}\bar{q}+q = \bar{p}+q$ ?  Is it absorption law ?

But absorption law is something different i knew. Please correct me.

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Distributive Property : $x+yz=(x+y)(x+z)$

$\bar p \bar q + q =(\bar p+q)(\bar q+q)$

Answer 2

Simple Trick to solve such questions without anything......

See Try to prove that the given option is not tautology by making T--->F i.e., Try to make LHS of Implication True and RHS be False...

Opt a)

((p v q)^(r v s)) -> ((p^r) v q vs)

See LHS can be true only when and is true 

p V q can be true either q is true or p or both.

if q is made true then RHS also becomes true as q is there with Or

So if p is made true then r V s of LHS has to be true. ie. r or  s has to be true to make LHS true. (so i take p to be true)

If s is made true  then RHS also becomes true which we do not want as T-->T is True

if r is made true and since we had taken p as true as LHS then RHS also becomes true thus there is no scope that RHS becomes False.thus it is tautology..

Now Opt b)

((p v q)^(r v s)) -> (q v s)

proceed as above.

 Try to make LHS T and RHS F.

If q on LHS is made true then RHS becomes True which we do not want

So p has to be true in order to make LHS "and" True.

Now Either r or s has to be true so if we take s true then LHS becomes true  and consequently RHS becomes true. but now we put r as true it becomes that T--->F so it is not tautology...

So you can proceed with other options...With Practice it becomes clear

Answer 3

Only (2) is Not Tautology