I'm starting to take a course in logic and the definition they gave us for a logical proposition is that it is a statement that is either true or false, but not both. My understanding is that "$x = 1$" is not a proposition, cause it depends on the value $x$ takes, so it could be true or false but we can't know which it is.

But looking at the statement "if $x = 1$, then $x + 1 = 5$" I could think maybe it is not a proposition because its hypothesis is not a proposition. On the other hand often in mathematics, this could be treated as a proposition, and we assume "$x = 1$" is true and work forward to see if "$x + 1 = 5$", which is not the case so the implication would be false.

But is it valid to define the truth-value of the whole statement without considering that "$x = 1$" could be false? In that case, the implication would be in fact true and thus we get (using the above result) it is true or false, but which is it? Can we define a unique truth-value? Is it right to say it has one and that it is a proposition?

The same situation arises for "if $x = z$, then $x + y = y + z$" (here I think the issue would be we don't have quantifiers nor a domain for $x$, $y$, and $z$).


2 Answers 2

  • The formula $$x=1$$ is variously called a predicate, a propositional function, and an open formula. You're right about this formula not being a proposition: a predicate typically has a varying truth value. (The predicate $x=x$ is definitely true though, and is thus a (logical) validity.)

    Similarly, $$\text{if }x = 1,\text{ then }x + 1 = 5\tag{*}$$ is also a predicate. It has the opposite truth value as the first predicate above.

  • On the other hand, the formula $$\text{for each $x, \, \big($if }x = 1,\text{ then }x + 1 = 5\big)\tag{#}$$ is a (false) proposition, also called a sentence and a closed formula. As you've pointed out, (in each interpretation) every proposition has a definite truth value.

What distinguishes a predicate from a proposition is whether the formula contains any free variable, that is, whether each variable in it is quantified by either $\forall$ or $\exists.$ (As illustrated above, a formula that has a fixed truth value is not necessarily a proposition!)

In practice though, implications in mathematical writing are typically implicitly universally quantified. This is the real issue in the given example: the predicate $(*)$ is actually intended to be understood as the proposition $(\#).$

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    $\begingroup$ Thanks a lot for your answer, it was very useful. PD : very good explanation. $\endgroup$
    – MMMagician
    Aug 25, 2021 at 1:53

A proposition is a statement that is "true" or "false" The statement "x= 1" is not a "proposition" because whether it is true or false depending on what x is.

But there is not a condition that the hypothesis and conclusion of a proposition, separately, be propositions. The statement "if x= 1 then x+ 1= 5" is a proposition because it is false.

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    $\begingroup$ "The statement 'if x= 1 then x+ 1= 5' is a proposition because it is false." I disagree: It's only false if x is 1, and vacuously true otherwise. But x is still not bound, so it's not a proposition. (Assuming we take "if ... then ..." to mean implication.) $\endgroup$ Aug 25, 2021 at 0:14

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