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Consider the following two statements.

• $S_1$: If a candidate is known to be corrupt, then he will not be elected
• $S_2$: If a candidate is kind, he will be elected

Which one of the following statements follows from $S_1$ and $S_2$ as per sound inference rules of logic?

1. If a person is known to be corrupt, he is kind
2. If a person is not known to be corrupt, he is not kind
3. If a person is kind, he is not known to be corrupt
4. If a person is not kind, he is not known to be corrupt

### 1 comment

Option C is correct.

S1:  P-->Q

S2:  X-->Z

If X is true then Z has to be true in order for statement S2 to be true.

If Z is true then Q is false. Now, if Q is false then P has to be false in order for statement S1 to be true.

\begin{align*} S_1 &= C \rightarrow \neg E\\ S_2 &= K \rightarrow E\\ \end{align*}

so, writing them using primary operators :
\begin{align*} S_1 &= \neg C \vee \neg E\\ S_2 &= \neg K \vee E\\ \end{align*}

on using resolution principle
$\neg E$ and $E$ cancels each other out
and conclusion = $\neg C \vee \neg K$

which can also be written as $K \rightarrow \neg C$ which is translated into English as = option C

Why option b is not the answer?
Because if person is not corrupt then he may or may not be kind.

See here if person is kind he is not corrupt . But if he is not corrupt then we cant say whether he is kind or not.
this might be useful to understand resolution principle

Option c. If a person is kind, he is not known to be corrupt

Let

• $C(x): x \text{ is known to be corrupt}$
• $K(x): x \text{ is kind}$
• $E(x): x \text{ will be elected}$

• $S1: C(x) \to \neg E(x)$
• $S2: K(x) \to E(x)$

S1 can be written as $E(x) \to \neg C(x)$ as $A \to B = \neg B \to \neg A$.
Thus, from S1 and S2,

$K(x) \to E(x) \to \neg C(x)$.

Thus we get C option.

@anoop
best explanation
Very good

Method for these kinds of question.

Use inference law : Here we use

Contrapositive law i.e. A $\rightarrow$ B is true then ~B $\rightarrow$ ~A is always true.

For  A $\leftrightarrow$ B is true the contrapositive true. converse(B$\leftrightarrow$ A) true and Inverse(~A$\leftrightarrow$~B)  also true.

Let ,

K: Person is kind

C: Person is corrupt

E: Person is elect

Here both statements 1,2 are premises and we need to check what is the conclusion.

S1: C-->¬E

S2: K-->E

3.  E-->¬C  from S1, and Contrapositive rule.

4. K-->¬C  from S2,3 and Hypothetical syllogism.It is valid.

K-->¬C ≡ " If a person is kind, he is not known to be corrupt ".

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