+5 votes
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Choose the correct alternatives (More than one may be correct).

Indicate which of the following well-formed formulae are valid:

1. $\left(P\Rightarrow Q\right) {\wedge} \left(Q \Rightarrow R\right) \Rightarrow \left(P \Rightarrow R\right)$
2. $\left(P\Rightarrow Q\right) \Rightarrow \left( \neg P \Rightarrow \neg Q\right)$
3. $\left(P{\wedge} \left(\neg P \vee \neg Q\right)\right) \Rightarrow Q$
4. $\left(P \Rightarrow R\right) \vee \left(Q \Rightarrow R\right) \Rightarrow \left(\left(P \vee Q \right) \Rightarrow R\right)$
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A is true it is Hypothetical syllogism.

In classical logic, hypothetical syllogism is a valid argument form which is a syllogism having a conditional statement for one or both of its premises. If I do not wake up, then I cannot go to work. If I cannot go to work, then I will not get paid. Therefore, if I do not wake up, then I will not get paid.

[(P⇒Q)∧(Q⇒R)]      =====>      (P⇒R)

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

T                                         F

You Can Not find   T====>F

So this formula is always true

## 3 Answers

+7 votes
Best answer

Ans) A

For reason follow pics

answered by Loyal (3k points)
selected by
+7 votes
To prove any wff valid or tautology try to use this analogy.

Since implication A->B is FALSE only when A=T and B=F.So to prove any implication is valid or not try to get TRUE->FALSE if we succeed then it is not valid,if we not then wff is valid.

So for option A

substitute P=T and R=F

RHS P->R become FALSE

LHS (P->Q)^(P->R)

To get true here we need T^T so substitute Q=T which makes P->Q TRUE and P->R FALSE so T^F=F which makes LHS=FALSE.

Hence we are unable to get T->F which proves wff given in OPTION A is valid.

NOTE: we can use similar kind of logic to prove contradiction.
answered by Boss (9.3k points)
LHS is (P->Q)^(P->R) or  (P->Q)^(Q->R)
+1 vote
Create a logic table for each of the given formula and check if we are getting true everywhere.
answered by Veteran (18k points)
Answer:

+3 votes
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