How do we decide when a statement's truth value depends in True/False question?

In my experience, when it is in a True/False question, then the answer is False. For example,

If $P(A)=0$, then $A =\emptyset$.

However, I choose False because (1) In general it is False and (2) the reasoning/implication is False since $P(A)=0$ does not necessarily imply $A =\emptyset$.

If we just leave the reasoning aside, only consider it as a statement. We know $P(A)=0$, it does not tell anything about the element in $A$. The set $A$ might contain some elements that have 0 probability, $A$ might indeed is empty. So it is like just asking you $A$ is empty or not without any helpful information. In such a case, do we say this statement is False?

An simple example can be:

If $x \in R $, then $x>0$.

Hence, if the statement in general is False but might be true in some case (i.e. depends), do we say it is False?

The problem arises when we say it is false while the negation is also False: $x \in R $ and $x \le 0$. So is it better to say it depends?

  • $\begingroup$ What does it mean "a statement's truth value depends" ??? In a given "context" a statement has a truth value: "now in my town it does not rain"; this is a fact. "Now I'm weraing a red shirt" is false, and so on. $\endgroup$ – Mauro ALLEGRANZA Jul 4 '17 at 19:24
  • $\begingroup$ We can consider three-valued logic to mangae "vague" situation. See also Vagueness. $\endgroup$ – Mauro ALLEGRANZA Jul 4 '17 at 19:26
  • $\begingroup$ @MauroALLEGRANZA "depends" means in the given "context", we cannot decide the truth value, and I provided 2 examples for it. $\endgroup$ – tautology Jul 4 '17 at 20:04
  • $\begingroup$ The answer below answer it; the truth-value of: "if $x \in \mathbb R$, then $x > 0$" depends on the value assigned to $x$ (in the domain $\mathbb R$). If we assign to $x$ the value $1$ the resulting "interpreted" formula is true, while with $0$ assigned to $x$ the formula is false. $\endgroup$ – Mauro ALLEGRANZA Jul 4 '17 at 20:08
  • $\begingroup$ @MauroALLEGRANZA (1) Thx for answering the question you asked in the comment, "What does it mean "a statement's truth value depends" ??? ". (2) And even better, that is what I am thinking, some statements cannot be decided to be true or false because it depends. $\endgroup$ – tautology Jul 4 '17 at 20:52

The "technical" term in logic is: satisfaction (to be true in an interpretation).

A sentence like $\forall x \ (x \ge 0)$ is satisfied in an interpretation with domain the set $\mathbb N$ of natural numbers; in symbols:

$\mathbb N \vDash \forall x \ (x \ge 0)$.

The same formula is not satisfied in an interpretation with domain the set $\mathbb R$ of real numbers; in symbols:

$\mathbb R \nvDash \forall x \ (x \ge 0)$.

In a specific interpretation $\mathcal I$, if a sentence $\sigma$ is true, then its negation $\lnot \sigma$ is false (see Law of excluded middle).

Things are different for "open" formulas, like $x > 0$; in this case we need a way to assign a "meaning" to the variable $x$ in a specific context.

If we consider an interpretation with domain $\mathbb N$ and we assign to $x$ as "temporary meaning" (there is a "technical" way to do this in logic, through variable assignment functions) the number $1$, we have that the formula $(x > 0)$ is satisfied: $\mathbb N \vDash (x > 0)[x \leftarrow 1]$.

If instead we assign to $x$ the value $0$, the formula is not satisfied.

Thus, in conclusion, the truth value in a given interpretation of a formula with a free variable, like $(x > 0)$, depends on the value assigned to $x$.

  • $\begingroup$ I am sorry that you misinterpret my question. You answer is totally correct, but I am asking how to decide T/F when a statement can be true/false in the given context, ie: depends. However, since it is a T/F question, we can only answer T or F. In the 2nd example, it is False, but the negation is also False. $\endgroup$ – tautology Jul 4 '17 at 20:14
  • $\begingroup$ @tautology - No; without a value assigned to $x$ the formula $x > 0$ is neither true nor false. A free variable in a math formula is like a pronoun in natural language: the phrase "it is red" has no defined meaning (or truth value) if we do not define what is the reference of "it". $\endgroup$ – Mauro ALLEGRANZA Jul 4 '17 at 20:19

I'll venture a wild guess: you're talking about "True/False" questions that some math instructors (myself included) like to ask on tests. There are several different situations that can appear on such a question.

  1. The question is very concrete, without any free variables. That's an easy case: "$2\times2=4$" is clearly True, while "$2\times2=5$" is clearly False.
  2. The question does contain some free variables in it, just like in both examples you provided. Two things about such questions: first of all, there's usually some context that's assumed to be clear both to the teacher and to the student; and second, they are silently, by default, assumed to be general "for all" kinds of statements, i.e. whether it is always true.

Turning to your examples.

(1) By asking whether

If $P(A)=0$, then $A=\emptyset$

is True or False, your instructor actually means to ask (roughly speaking) whether

For any set $A$, if $P(A)=0$, then $A=\emptyset$

is True or False. Since, as you correctly explained, this is not always true, the correct answer is False.

(2) Similarly, by asking whether

If $x\in\mathbb{R}$, then $x>0$

is True or False, your instructor actually means to ask (roughly speaking) whether

For any $x$, if $x\in\mathbb{R}$, then $x>0$

is True or False. Again, this is False because some real numbers are not positive (even though some are).

Short summary: the usual interpretation of such questions is that:

  1. True means ALWAYS True;

  2. while anything else, i.e. it depends because sometimes it's true and sometimes it's false, is treated as False.

Sorry for such a long answer, but I should also address two misconceptions in your post.

(1) The answer is not always False. For example:

If $\det A\neq0$, then $A$ has an inverse matrix.

I intentionally made it sloppy, but that's how it often is. On the one hand, we should specify that $A$ is a square matrix so that the question is meaningful. On the other hand, the very fact of using "$\det A$" assumes that $A$ is a square matrix. Mathematically, we can get very nitpicky about this. But in the reality of college math tests, these statements are pretty much always meaningful; the only question is whether they are true or false.

(2) The negation of "If $A$, then $B$" is NOT "If $A$, then not $B$". So your reasoning in the last paragraph of your post is wrong. There's nothing wrong with both "if $x\in\mathbb{R}$, then $x>0$" and "if $x\in\mathbb{R}$, then $x\le0$" being False because they are NOT negations of each other.

  • $\begingroup$ Sorry I don't get it for your example "det A". Seems like it is true anyway. Like what you said, when we applied det for A, A must be a square matrix. There is no ambiguity. And Thanks for pointing it out for the negation, I meant to say "A and not B", already changed it in the question $\endgroup$ – tautology Jul 4 '17 at 20:44
  • $\begingroup$ And to be clear, when people phrases "if $x \in R$", doesn't they refer to "Let $x \in R$" or simply "$x \in R$"? It just states what is x. So in this very example, if we are sloppy, we can say the negation is "if $x \in R$, then $x \le 0$", which is interpreted as "Let $x \in R, x \le 0$" $\endgroup$ – tautology Jul 4 '17 at 22:42
  • $\begingroup$ @tautology: You can interpret the original statement like that... but you can NOT negate it like that. As Mauro ALLEGRANZA already explained, a statement like "$x>0$" does NOT have a truth value because $x$ is undefined in it. In fact, it's not even a statement -- it's a proposition with a variable $x$. $\endgroup$ – zipirovich Jul 7 '17 at 10:52
  • $\begingroup$ @tautology: The actual original statement is not "$x>0$", nor even "If $x\in\mathbb{R}$, then $x>0$" -- they both are propositions (with a variable $x$), not statements. The actual statement (in the sense that this is what the instructor actually meant) is "$\forall x$, If $x\in\mathbb{R}$, then $x>0$". And its negation is "$\exists x$ such that $x\in\mathbb{R}$ and $x\le0$". Now it should be clear that the original statement is false and its negation is true. No fuzzy logic or anything weird here! $\endgroup$ – zipirovich Jul 7 '17 at 11:01

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