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"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

Options are $:$

  1. $\forall x(P(x) \implies (G(x) \wedge S(x)))$
  2. $\forall x((G(x) \wedge S(x)) \implies P(x))$
  3. $\exists x((G(x) \wedge S(x)) \implies P(x))$
  4. $\forall x((G(x) \vee S(x)) \implies P(x))$

I try to explain $:$ At the same time an ornament can not be gold and silver , it should be either gold are silver , therefore option $(4)$ is correct .


Please check whether my solution is correct ?

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  • $\begingroup$ So what's the question? $\endgroup$ – skyking Oct 8 '15 at 8:57
  • $\begingroup$ Please check whether my solution is correct ? $\endgroup$ – user278174 Oct 8 '15 at 9:05
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You are correct .

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

$\forall x((G(x) \vee S(x)) \implies P(x))$

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Yes, the interpretation of "and" in natural language(s that I know) is that between nouns or pronouns they should be interpreted as the conjunction of the statements formed by keeping each noun/pronoun.

For example this means that "Gold and silver ornaments are precious" should be interpreted as $\textrm{"Gold ornaments are precious"} \land \textrm{"Silver ornaments are precious"}$.

Furthermore "Gold ornaments are precious" means that if something is a "Gold ornament" then it's "Precious" (and as not qualified this is supposed to hold universally).

This boils down to it being the same as $(\forall x G(x)\rightarrow P(x)) \land (\forall x S(x)\rightarrow P(x))$ which is equivalent to 4.

That 4 is not equivalent to any of the other is simple predicate calculus that's left as an excercise.

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  • $\begingroup$ Thanks for your explanation. $\endgroup$ – user278174 Oct 8 '15 at 9:28

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