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.

Dark Mode

6,628 views

28 votes

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?

- If a person is known to be corrupt, he is kind
- If a person is not known to be corrupt, he is not kind
- If a person is kind, he is not known to be corrupt
- If a person is not kind, he is not known to be corrupt

33 votes

Best answer

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

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**

63 votes

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.

10 votes

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.

4 votes

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 ".**