Question

Assume that a Creative (\(C\)) person will Succeed (\(S\)) if the person is also Disciplined (\(D\)), but will not succeed otherwise. Now, consider the following statements:
(i) \(C \wedge S \Leftrightarrow D\)
(ii) \(C \Rightarrow (S \Leftrightarrow D)\)
(iii) \(C \Leftrightarrow ((D \Rightarrow S) \vee \neg S)\)
Which of the following options is correct?

• A. Both (i) and (ii) are TRUE
• B. Only (ii) is TRUE
• C. Both (ii) and (iii) are TRUE
• D. Only (iii) is TRUE

Hint:

Rewrite the given rule as C => (S <=> D), then test each statement against the valid (C,S,D) combinations.

Answer 1

The Correct Option is B

Step 1: Translate the given rule into logic.
The rule says: if a person is Creative, then they succeed exactly when they are also Disciplined. In symbols, this is \(C \Rightarrow (S \Leftrightarrow D)\), which is precisely statement (ii). So statement (ii) is just a direct, correct restatement of the given rule, and must be TRUE.

Step 2: List the valid combinations of \(C, S, D\) allowed by the rule.
When \(C\) is false, the rule places no restriction, so \(S\) and \(D\) can be anything: this gives 4 valid combinations \((C{=}0,S{=}0,D{=}0)\), \((0,0,1)\), \((0,1,0)\), \((0,1,1)\).
When \(C\) is true, we need \(S \Leftrightarrow D\), so \(S\) and \(D\) must match: this gives 2 more valid combinations \((C{=}1,S{=}0,D{=}0)\) and \((1,1,1)\).
In total there are 6 valid worlds consistent with the rule.

Step 3: Test statement (i), \(C \wedge S \Leftrightarrow D\), against these worlds.
Take the valid world \((C{=}0, S{=}0, D{=}1)\). Here \(C \wedge S = 0\) but \(D = 1\), so \(C \wedge S \Leftrightarrow D\) evaluates to false. Since (i) fails in a world where the original rule still holds, (i) is not always TRUE, so (i) is FALSE overall.

Step 4: Test statement (iii), \(C \Leftrightarrow ((D \Rightarrow S) \vee \neg S)\), against these worlds.
Take the valid world \((C{=}0, S{=}0, D{=}0)\). Here \(D \Rightarrow S\) is \(0 \Rightarrow 0 = 1\), and \(\neg S = 1\), so the right side is \(1 \vee 1 = 1\). But \(C = 0\), so \(C \Leftrightarrow 1\) is false. Since (iii) fails in a valid world, (iii) is not always TRUE, so (iii) is FALSE overall.

Step 5: Conclude.
Only statement (ii) holds in every world consistent with the given rule; (i) and (iii) each fail in at least one such world.

Final Answer:
Only (ii) is TRUE, which is option (B).