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I am trying to correctly use predicate symbols and using the appropriate quantifiers were I have to write each English language statement in predicate logic and the domain is the whole word.

$P(x)$ is "$x$ is a person."

$T(x)$ is "$x$ is a time."

$F(x_1,x_2)$ is "$x_1$ is fooled at $x_2$."

$1.$ You can fool some of the people all of the time.

My Answer at best: $\exists x(P(x) \rightarrow \forall y(T(y) \rightarrow F(x_1,x_2))$

$2.$ You can fool all of the people some of the time.

My answer at best: $\forall x(P(x) \rightarrow \exists y(T(y) \rightarrow F(x_1,x_2))$

$3.$ You can’t fool all of the people all of the time.

My answer at best: $\neg (\forall x(P(x) \rightarrow \forall y(T(y) \rightarrow F(x_1,x_2)))$

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  • $\begingroup$ I can't be sure about the ∀ and the ∃ in the right places $\endgroup$ – Patrick William McCully Apr 21 '15 at 3:52
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  1. $$\exists x \exists y (P(x) \wedge P(y) \wedge \forall z T(z) \wedge F(x,y)) $$

  2. $$ \exists x (P(x) \wedge \exists z T(z) \wedge (\forall y P(y) \rightarrow F(x,y)) ) $$

  3. $$ \neg ( \exists x (P(x) \wedge ((\forall z T(z) \wedge \forall y P(y) ) \rightarrow F(x,y)) )) $$

In you answer, $x_1$ and $x_2$ are free variables. Make sure all variables are bounded.

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  • $\begingroup$ I just notice there is P(x), T(x), and F(x1,x2). There probably wouldn't be any y or z, just only x, x1, and x2. For example for number 1 would be $\exists x \exists x (P(x) \wedge P(x) \wedge \forall x T(x) \wedge F(x1,x2))$ maybe? $\endgroup$ – Patrick William McCully Apr 21 '15 at 4:17
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    $\begingroup$ just like x, y and z are variables. I strongly encourage you to read the book amazon.com/The-Logic-Book-Merrie-Bergmann/dp/0078038413 $\endgroup$ – Proton Boss Apr 21 '15 at 4:20

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