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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
in Mathematical Logic
edited by
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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.

4 Answers

23 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


edited by
0
@amarVashishth

How do we apply resolution principle ?
2
@junaid

If you have disjunction of literals given in the premises you can cut out the literals which are in true as well as complemented form in two different premises.
53 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.

1
@anoop  
best explanation
1
@anoop Best answer
0
Very good
7 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.


edited by
2 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 ".

So, (c) is the Ans.

Answer:

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