GATE CSE 2015 | Set 2 | Question: 3

Question

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?

• A. If a person is known to be corrupt, he is kind
• B. If a person is not known to be corrupt, he is not kind
• C. If a person is kind, he is not known to be corrupt
• D. If a person is not kind, he is not known to be corrupt

Comments

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.

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Detailed Video Solution: https://youtu.be/B6xo-2A8ano?t=870&feature=shared

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Propositional Logic Complete Summary & ALL GATE PYQs: https://youtube.com/playlist?list=PLgjejdknTfWNwpjPp4nYfxPbj5fqMeOsR&feature=shared

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• Mark My word ------>> "IMPLICATION IS TRANSITIVE" So if you apply this in this question it will be simple.....

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c: A Candidate is known to be corrupt.
e: He will be elected.
k: A candidate is kind.

S1: c --> ~e        Contrapositive of S1: e --> ~c
S2: k --> e         Contrapositive of S2:~e --> ~k

By Transitivity of Implications, 
Considering S1 and contrapositive of S2: c --> ~e --> ~k 
If a candidate is known to be corrupt, he is not kind.
Considering S2 and contrapositive of S1: k --> e --> ~c
If a candidate is kind, he is not known to be corrupt.

Only option C matches. 
Hence, Answer:C

Answer 1

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

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

Comments

Why option b is not the answer?

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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.

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got it

Answer 2

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.

Comments

@anoop  
best explanation

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@anoop Best answer

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Very good

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got it , simple

Answer 3

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.

Answer 4

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 5

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Comments

First we will convert S1 and S2 into logical forms:

   S1: C --> ~E

   S2: K --> E

We will assume inference to be false and try to make premises S1 and S2 true , if we can make S1 and S2 to be true then given inference is not valid.

a) C --> K

 Let C -->K = False(F) , then C is true and K is false.

Now, S1: C --> ~E

   C=true, ~E=true, thus S1 is true.

Now, S2: K --> E

   K=false, E=false, thus S2 is true.

Since S1 and S2 are both true thus it is not valid argument.

b) ~C --> ~K

 Let ~C --> ~K = False(F) , then C is false and K is true.

Now, S1: C --> ~E

   C=false, ~E=false, thus S1 is true.

Now, S2: K --> E

   K=true, E=true, thus S2 is true.

Since S1 and S2 are both true thus it is not valid argument.

c) K --> ~C

 Let K --> ~C = False(F) , then K is true and C is true.

Now, S1: C --> ~E

   C=true, ~E=true, thus S1 is true.

Now, S2: K --> E

   K=true, E=false, thus S2 is false.

Since S2 is false thus it is valid argument.

d) ~K --> ~C

 Let ~K --> ~C = False(F) , then K is false and C is true.

Now, S1: C --> ~E

   C=true, ~E=true, thus S1 is true.

Now, S2: K --> E

   K=false, E=false, thus S2 is true.

Since S1 and S2 are both true thus it is not valid argument.

 

Thus option (c) is correct.

Answer 6

this might be useful to understand resolution principle