Prepositional logic :1.1

Question

Derive clause (P ∨ Q) from clauses (P ∨ R), (Q ∨ ¬R)
is following statements related to this rule is FALSE?

(P ∨ Q) ⇒ FALSE if and only if both P and Q are unsatisfiable

Comments

P V Q

Q V $\sim R$

--------------------------

P V Q.

Simplified as $[(PVQ)\wedge (QV\sim R)] ->PVQ$.

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No thiz is not disjunction syllogism

check again

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To derive any clause from two statements we use conjuction.

For instance 

P->R is derived from P->Q and Q->R.

[(P->Q) $\wedge (Q->R)$]->P->R.

To check above is right or wrong.

Try to make right side part as F.//it is possible only when P=T and R=F.

Try to make left part as TRUE if it is above predicate is wrong,otherwise it is correct.

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[(P->Q) ∧(Q−>R)]->(P->R) thiz is tautology , hypothetical syllogism propertry but

sorry , but i m not getting what u want to say .

 my Question is (P ∨ Q) ⇒ FALSE if and only if both P and Q are unsatisfiable .

Acutally i m not getting thiz question which i asked ...thiz is previous yr ques .

Note: ya if P and Q are false then it leads to whole thing true . due to p -> q = p ' + q

can u tell what is question ?

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what is ask to prove ?

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P V Q->FALSE.

i think if both values are false then only the whole predicate is true.remaining all cases it is false.

If P=T then P V Q=T and T->F is false.

this is my understanding,what u think??

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ya tht right .  It is previous yr gate question .

What type of english they use i don't get yr .What they want to prove It true or false ?

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@santosh

c thiz

https://gateoverflow.in/99007/propositional-logic-converse

i think if exam is gate thn it leads to false .

i m laughting still 3 minutes . lol :)