# Can a predicate be a contradiction?

In the university course I'm taking, a predicate is defined as a mathematical statement whose truth value depends on the variables involved in the statement. This definition makes me wonder whether a predicate can be a contradiction.

For instance, if we use $$P(x)$$ to denote the statement "$$x$$ is an even number", clearly $$P(x)$$ itself is a predicate.

However, if we consider the statement

$$x$$ is an even number and $$x$$ is not an even number,

it can be written as $$P(x)\land\lnot P(x)$$.

I think based on my knowledge it is a valid statement, but I am not sure whether it is a predicate. The part I am not sure about is:

1. By the law of excluded middle and well-definedness of mathematical statements, the statement should always be false regardless of the value of $$x$$. Hence, without knowing the value of $$x$$, we can say that this statement is false. In this sense, the truth value of this statement does not depend on the value of $$x$$, so it should not be a predicate.

2. However, I feel in the above argument (1), I unconsciously use $$\forall$$ to quantifier the statement, which makes it a proposition (or at least we can directly tell the truth value without considering $$x$$). If this is true, this statement should be a predicate.

I feel I could not figure out which of my thoughts is true, or both of them are wrong, could someone please help me on this?

• I would say that a "predicate" is expressed by a predicate symbol $P(x)$. A formula with a free variable $\varphi(x)$ expresses a "property". Dec 1, 2021 at 8:16
• If we follow this approach, we may have the formula $\varphi(x) := Px \land \lnot Px$ that "defines" the "empty" property, i.e. the property whose extension is the empty set. Dec 1, 2021 at 9:35

$$P(x)\land\lnot P(x).$$

I think based on my knowledge it is a valid statement, but I am not sure whether it is a predicate.

By definition, a predicate is a formula with a free variable.

$$P(x)$$ is typically understood to mean something like $$∃y\;x=2y,$$ but not $$∃x\;x=2x.$$ In other words, by convention, $$P(x)\land\lnot P(x)$$ does indicate a predicate.

I think based on my knowledge it is a valid statement

A statement typically means a formula with no free variable; the given formula is a “well-formed formula” rather than a statement.

It is clearly a contradiction—i.e., its truth-functional form is false regardless of interpretation—so cannot be valid—i.e., true regardless of interpretation—either.

In general, some predicates are unsatisfiable—false regardless of interpretation—some predicates are valid—true regardless of interpretation—some predicates are satisfiable but invalid.

1. By the law of excluded middle and well-definedness of mathematical statements, the statement should always be false regardless of the value of $$x$$. Hence, without knowing the value of $$x$$, we can say that this statement is false. In this sense, the truth value of this statement does not depend on the value of $$x$$, so it should not be a predicate.

2. However, I feel in the above argument (1), I unconsciously use $$\forall$$ to quantifier the statement, which makes it a proposition (or at least we can directly tell the truth value without considering $$x$$). If this is true, this statement should be a predicate.

This is unsound reasoning, based on the misconception that a predicate's truth value is necessarily dependent on its free variable. (Compare this characterisation with the definition in the first line.)

The fact that the given formula remains true upon quantification does not negate the fact that it is a predicate.

• Thanks! I get it now. But just to clarify myself, the word 'valid' was meant to indicate that I thought the sentence is indeed a statement, rather than the truth value of this statement. Dec 1, 2021 at 5:26
• @Doero I knew what you meant, and was highlighting that the technical meaning of 'valid' means that we have to be careful about using it in the layman sense. I occasionally make this slip too. Dec 1, 2021 at 5:27
• And also, from your response, I felt you made a distinction between statement and well-formed formula, is there a difference between them? From the construction of wff I feel they are the same in some sense. Dec 1, 2021 at 5:27
• @Doero Every statement (no free variable) is a formula (can have a free variable), but not vice versa. Dec 1, 2021 at 6:24
• @Doero P.S. When I find myself wanting to use 'valid' in a sense that contradicts its formal-logic meaning, I frequently say 'legit' instead. Dec 1, 2021 at 7:22

We can, with no problems whatsoever, define a unary predicate symbol $$Q$$, which is defined such that $$Q(x)$$ is equivalent to the statement "$$x$$ is even and $$x$$ is not even". And, as you've noted, this is the same as the statement $$\lnot T$$, i.e. it is a statement that is false regardless of the value of $$x$$, meaning that the formula $$\forall x \lnot Q(x)$$ is a true sentence.

This is fine, and it is analogous to the fact that we can define a constant function, i.e. $$f(x) = c$$. $$f$$ is a function, it operates on $$x$$, but the operation it performs gives the same result regardless of $$x$$, but we would still say that $$f$$ is a function of $$x$$.

Importantly, we would definitely regard $$Q$$ as being a predicate, and hence dependent (if somewhat vacuously) on $$x$$, so $$Q(x)$$ is not a sentence as it has a free variable.

To think of it another way, consider that a lot of the time we take a predicate that we think is true for all of its inputs, and then try to prove it as such. If we have a predicate $$P(x)$$ and are trying to prove $$\forall x P(x)$$, how could $$P$$ suddenly become a different type of object depending on whether we succeed or not?

• Thanks! The analogy to constant function is really helpful! Dec 1, 2021 at 5:24