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Consider the following statement:
$$\forall x(x > 0) \to (x < 2)$$

Obviously, this statement is false. We can easily contradict it; for example when $x=5$.

Now, consider this statement:
$$(\forall x(x > 0)) \to (x < 2)$$

As far as I understand, the $x$ in the left-hand side is bounded, while the $x$ in the righ-hand side is free. They're different variables. Yet, I don't understand why this statement is always true.

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It is true because what is on the left side of the implication is false (assuming your universe is all the real numbers). – Harald Hanche-Olsen Feb 2 '14 at 16:19
up vote 3 down vote accepted

The proposition $(\forall x(x > 0))$ is false. Therefore, it implies $(x < 2)$, since a false proposition implies anything ("When Hell freezes over; when pigs fly").

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Right, so simple. thanks! – SuperStamp Feb 2 '14 at 16:22

The second statement

$$\big(\forall x(x > 0)\big) \to (x < 2)$$

is not true (assuming this is a statement of the usual first-order logic)!

It's not false either, that is, you can't assign a true/false value to a ill-formed statement, it would be nonsensical.

I know we would like it to be true, and there is even a convention where free variables are bound with universal quantifiers, but formally it is not.

You could solve this problem by making this statement into a well-formed one by binding the free variable, or you can extend the system to allow valuations of such expressions. Then it could be true (perhaps it could be also false if we had extended the system in some weird way), but not before. In fact, even with $\Phi \equiv \mathtt{false}$ the formula $\big(\forall x\ (\Phi)\big) \to (5<2)$ would be false in an empty universe.

Beware of the free variables and don't get tricked $\ddot\smile$

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