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$P(x): x$ is precious
 
$P(x): x$ is precious
 
 
(A) $\forall;x(P(x) \implies (G(x) \wedge S(x)))$
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(A) $\forall x(P(x) \implies (G(x) \wedge S(x)))$
 
 
<b>(B) </b>&forall;x((G(x) &and; S(x)) &rarr; P(x))
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(B) $\forall x((G(x) \wedge S(x)) \implies P(x))$
<br>
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<b>(C) </b>&exist;;x((G(x) &and; S(x)) &rarr; P(x))
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(C) $\exists x((G(x) \wedge S(x)) \implies P(x))$
<br>
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(D) </b>&forall;x((G(x) &or; S(x)) &rarr; P(x))
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'''(D) $\forall x((G(x) &or; S(x)) \implies P(x))$'''
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
 
==={{Template:Author|Happy Mittal|{{mittalweb}} }}===
<b>Sol : </b> Basically statement is saying that for every thing, if it is a Gold ornament or a silver ornament, then it is precious.
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Basically statement is saying that for every thing, if it is a Gold ornament or a silver ornament, then it is precious.
<br>
+
So &forall;x((G(x) &or; S(x)) &rarr; P(x)) is correct logical formula, and therefore option <b>(D)</b> is correct.
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So, $\forall x((G(x) &or; S(x)) \implies P(x))$ is correct logical formula.
 
{{Template:FBD}}
 
{{Template:FBD}}
  
 
[[Category: GATE2009]]
 
[[Category: GATE2009]]
 
[[Category: Logical Inference questions]]
 
[[Category: Logical Inference questions]]

Revision as of 14:11, 14 July 2014

Which one of the following is the most appropriate logical formula to represent the statement? $``$Gold and silver ornaments are precious$$.

The following notations are used:

$G(x): x$ is a gold ornament

$S(x): x$ is a silver ornament

$P(x): x$ is precious

(A) $\forall x(P(x) \implies (G(x) \wedge S(x)))$

(B) $\forall x((G(x) \wedge S(x)) \implies P(x))$

(C) $\exists x((G(x) \wedge S(x)) \implies P(x))$

(D) $\forall x((G(x) ∨ S(x)) \implies P(x))$

Solution by Happy Mittal

Basically statement is saying that for every thing, if it is a Gold ornament or a silver ornament, then it is precious.

So, $\forall x((G(x) ∨ S(x)) \implies P(x))$ is correct logical formula.



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Which one of the following is the most appropriate logical formula to represent the statement? $``$Gold and silver ornaments are precious$$.

The following notations are used:

$G(x): x$ is a gold ornament

$S(x): x$ is a silver ornament

$P(x): x$ is precious

(A) $\forall;x(P(x) \implies (G(x) \wedge S(x)))$

(B) ∀x((G(x) ∧ S(x)) → P(x))
(C) ∃;x((G(x) ∧ S(x)) → P(x))
(D) </b>∀x((G(x) ∨ S(x)) → P(x))

Solution by Happy Mittal[edit]

Sol : Basically statement is saying that for every thing, if it is a Gold ornament or a silver ornament, then it is precious.
So ∀x((G(x) ∨ S(x)) → P(x)) is correct logical formula, and therefore option (D) is correct.



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