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Let $P(x),Q(x),R(x)$ be the statements $x$ is a clear explanation,$x$ is satisfactory,$x$ is an excuse,respectively. Suppose that the domain for $x$ consists of all the English text. Express each of these statements using quantifiers, logical connectives and $P(x),Q(x),R(x)$.

a. All clear explanations are satisfactory.

b. Some excuses are unsatisfactory.

c. Some excuses are not clear explanations

Please Corret Me

a. $\forall x ~ (P(x) \to Q(x))$

b. $\exists x ~ (R(x) \to \neg Q(x))$

c. $\exists x ~ (R(x) \to \neg P(x))$

I am not completely sure why I have used implication instead of a conjunction, I am wondering can someone explain,

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Except in the most obvious cases, it is probably better not to use implication. The reason is that the informal meaning of implication is not quite the same as the formal meaning of the logical connective, making mistakes more likely. –  André Nicolas Dec 13 '12 at 17:52
    
@AndréNicolas , For my answers in b,c (above) I tried to change implication to the disjunction and changing 'existence' to 'for all' and I find it very obvious that I am wrong , As Always,Thanks for the help. –  Hooman Dec 13 '12 at 18:04

3 Answers 3

up vote 2 down vote accepted

(b) and (c) are wrong. Restricted existentials need conjunctions. 'Some $A$s are $B$s" says that something which is an $A$ is also a $B$, so $\exists x(Ax \land Bx)$.

$\exists x(Ax \to Bx)$ is true if there is something which satisfies the condition $Ax \to Bx$, and anything that doesn't satisfy $A$ will make the antecedent of the condition false and hence make the condition true. So $\exists x(Ax \to Bx)$ is true if something isn't $A$, which isn't what you want at all.

Any elementary logic text will explain how to express restricted quantifications and stop you making this very elementary mistake in translation -- e.g. look for Paul Teller's excellent Primer, now freely available online.

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Assuming that by Vx you mean $\forall x$ (for all x) you are correct on the first one. The reason you use implication is that you are transcribing "all A are B" into "A(x) implies B(x)" which is an implication.

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It should be $\wedge$ in both b and c. It might be more intuitive to write (for example, b) $\exists x~ \neg (R(x) \to Q(x))$.

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