# Is 2 <= x <= 1 a predicate or a statement?

I can understand that a predicate is a predicate because the truth or falsity of it depends on the specific value of the 'variable' or the 'part acting as a variable' in it. (specific value should be from the domain of course).

But let's have this predicate: 2 <= x <= 1 (Domain is all real numbers)

Now whatever value we substitute for x, the predicate results in a statement which is false and hence there is absolutely no doubt about its truth value. It is false independent of the value replaced.

Does this make this predicate a statement essentially?

• @NoahSchweber Sorry! It was a typo. Edited. It is unconditionally false. – Parker Queen Oct 21 '18 at 17:01
• It is an open formula : $(2 \le x) \land (x \le 1)$ because the variable x is not quantified. – Mauro ALLEGRANZA Oct 21 '18 at 17:11
• In predicate logic, a sentence is "a well-formed formula with no free variables. A sentence can be viewed as expressing a proposition, something that must be true or false. The restriction of having no free variables is needed to make sure that sentences can have concrete, fixed truth values. As the free variables of a (general) formula can range over several values, the truth value of such a formula may vary." – Mauro ALLEGRANZA Oct 21 '18 at 17:14

I'd caution against defining a predicate as having a truth value "depending on" the arguments; one should say it's determined by them, which is consistent with all choices given the same result. It's a bit like functions. You presumably want the sum of two $$\mathbb{R}\mapsto \mathbb{R}$$ functions to also be a function, but $$x^3,\,1-x^3$$ have a constant sum. Similarly, you'd want the conjunction of $$2\le x,\,x\le 1$$, with $$x$$ fixed (variable), to be a statement (predicate), just like the conjuncts themselves are. So even through $$2\le x\le 1$$ is identically false, it's still a statement if $$x$$ is specified, or a unary predicate if it isn't.
A statement (sentence) does not have any free variables. So in isolation, $$2 \le x \le 1$$ is not a sentence, for syntactical reasons. If you add a quantifier it becomes a sentence: $$\forall x: 2 \le x \le 1$$. This is a false sentence in the model where the domain is all real numbers and the symbols have their usual meanings. Alternatively you can add an existential quantifier: $$\exists x: 2 \le x \le 1$$. This is obviously also a false sentence.
Viewed as a predicate, $$2 \le x \le 1$$ is a perfectly good predicate since it has $$x$$ as a parameter. Compare with a function $$f(x)$$ where a special case is $$f(x)=0$$ for all values of $$x$$. Here $$f$$ is still a function rather than a constant symbol, for syntactical reasons.