# Is this vacuously true? Is both true and vacuously true allowed?

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• Happiness
In summary, the statement "if ##x>-1## then ##f(x)<0##" is vacuously true for the given function, since the function is undefined at x=1. However, the statement can also be considered non-vacuously true for all values of x except 1. It is not possible for a statement to be both non-vacuously true and vacuously true at the same time. The existence of f(x) is not explicitly stated as part of the hypothesis, but it is implied. The statement "if ##x>-1## then ##f(x)<0##" is equivalent to "##f(x)<0## for all values of ##x>-1##". However, the statement

#### Happiness

Consider the statement "if ##x>-1## then ##f(x)<0##". Given that ##f(1)## is undefined, is the statement true or false or vacuously true for the following function?

##f(x)=\begin{cases}\frac{1}{x^2-1},&x\leq1\\ \frac{1}{1-x^2},&x>1\end{cases}##

If it's true, would you say it is non-vacuously true for all values of ##x## except ##1## and vacuously true for the case ##x=1##? Can a statement be both non-vacuously true and vacuously true at the same time?

A conditional statement with a false antecedent is vacuously true. But is the existence of ##f(x)## part of the antecedent?

Happiness said:
Consider the statement "if ##x>-1## then ##f(x)<0##". Given that ##f(1)## is undefined, is the statement true or false or vacuously true for the following function?

##f(x)=\begin{cases}\frac{1}{x^2-1},&x\leq1\\ \frac{1}{1-x^2},&x>1\end{cases}##
The definition of f above is erroneous, as f is undefined when x = 1.

I believe there is a typo in the function definition. As written, the statement ##if x > -1##, then ##f(x) < 0## is false, since f is not defined at every value in the interval ##(-1, \infty)##.
Happiness said:
If it's true, would you say it is non-vacuously true for all values of ##x## except ##1## and vacuously true for the case ##x=1##? Can a statement be both non-vacuously true and vacuously true at the same time?

A conditional statement with a false antecedent is vacuously true. But is the existence of ##f(x)## part of the antecedent?
How do you interpret the antecedent as being false? The hypothesis here is x > - 1.

Mark44 said:
How do you interpret the antecedent as being false? The hypothesis here is x > - 1.

I am not sure if "##f(x)## exists" is part of the hypothesis.

For example, "all cell phones in the room are turned off" is true whenever there are no cell phones in the room because "a cell phone exists in the room" is part of the hypothesis even though it is not explicitly stated as such. To state it explicitly, the statement would become "if a cell phone exists in the room, then all cell phones in the room are turned off".

So is the statement "if ##x>-1## then ##f(x)<0##" equivalent to the statement "if ##x>-1## and if ##f(x)## exists, then ##f(x)<0##"?

Happiness said:
I am not sure if "##f(x)## exists" is part of the hypothesis.

For example, "all cell phones in the room are turned off" is true whenever there are no cell phones in the room because "a cell phone exists in the room" is part of the hypothesis even though it is not explicitly stated as such. To state it explicitly, the statement would become "if a cell phone exists in the room, then all cell phones in the room are turned off".

So is the statement "if ##x>-1## then ##f(x)<0##" equivalent to the statement "if ##x>-1## and if ##f(x)## exists, then ##f(x)<0##"?
That's more or less how I read it, with a slight modification.
If x > -1, and f is defined as shown in the piecewise definition, then f(x) < 0.
The "and f is defined as ..." part is implied.
The stuff between "if" and "then" is the hypothesis, and the conclusion is f(x) < 0.

What sort of class or textbook is this from? If it's for an ordinary math class, I still maintain that there's a typo in the problem. If the class or textbook is more about logic and implications, then I would say that the statement is vacuously true -- the hypothesis is false (since f is not defined at every point in the interval of interest), which make the conclusion irrelevant.

Happiness
There are many values of x > -1 where f(x) exists.

mfb said:
There are many values of x > -1 where f(x) exists.

"If ##x>-1## then ##f(x)<0##" is equivalent to "##f(x)<0## for all values of ##x>-1##.

The statement ##x>-1## then ##f(x)<0##" is vacuously true, I believe, because its explicit meaning is "for all values of ##x>-1##, if ##f(x)## exists, then ##f(x)<0##. But since ##f(x)## does not exist for all values of ##x>-1##, the hypothesis is not satisfied, and hence it's vacuously true.

"There exists a value of ##x>-1## such that ##f(x)## exists and ##f(x)<0##" is non-vacuously true. But this is a different statement from "if ##x>-1## then ##f(x)<0##".

But the statement "for all values of ##x>-1, f(x)## exists and ##f(x)<0##" is false.

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Happiness said:
The statement ##x>-1## then ##f(x)<0##" is vacuously true, I believe, because its explicit meaning is "for all values of ##x>-1##, if ##f(x)## exists, then ##f(x)<0##. But since ##f(x)## does not exist for all values of ##x>-1##, the hypothesis is not satisfied, and hence it's vacuously true.
That's not how the logic works. And even if it would, it would not be vacuously true because there are x values.
What does x>-1 even mean if we don't start with defining what x can be? x would not even have to be a real number. "For all x in the domain of f where x>-1, f(x)<0."
The domain of the function is probably R\{1}, adding the constraint x>-1 we get the union of two intervals, [-1,1) u (1,inf). The function is defined for all x out of it by construction.

But the statement "for all values of ##x>-1, f(x)## exists and ##f(x)<0##" is false.
It is not even well-defined because you didn't specify what x is, and how we can compare it to -1.

mfb said:
What does x>-1 even mean if we don't start with defining what x can be? x would not even have to be a real number.

For the statement, ##x## is a real number, not necessarily in the domain of ##f(x)##.

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It is necessarily the domain of f, this is not always written explicitly but always implied. If it would not be, why should it be a real number at all? Why not a 2x3 matrix? There is no such thing as "the domain of f(x)", as f(x) is a number, not a function.

mfb said:
It is necessarily the domain of f, this is not always written explicitly but always implied. If it would not be, why should it be a real number at all? Why not a 2x3 matrix? There is no such thing as "the domain of f(x)", as f(x) is a number, not a function.

This is a question on logic. Suppose ##x## is a real number. Is the statement "if ##x>-1## then ##f(x)<0##" true or false?

"Three is more than two" is a true statement.

Q1. Is the statement "an apple is more than an orange" non-vacuously true or false or vacuously true?

Obviously, such a statement doesn't make sense. But I want to find out how do we go about determining whether a statement is vacuously true or not. Consider the statement "all cell phones in the room are turned off". When there are no cell phones in the room, it doesn't make sense, at least in normal everyday usage. But in this case, the statement is vacuously true.

Q2. How about the statement "all cell phones in the room are turned xaxana"? (Given that a cell phone is either turned on or turned off but never both and never neither.)

Obviously, "xaxana" is undefined and doesn't make sense. But a statement is either true or false. There is no third classification called nonsense.

Q3. All cell phones in the xaxana are turned off.

Q4. All xanana in the room are turned off.

I think Q3 and Q4 are both vacuously true because the premise "there is a cell phone in the xaxana" and the premise "there is a xaxana in the room" cannot be satisfied. For the same reason, Q2 is vacuously true if there are no cell phones in the room. If there is at least one cell phone in the room, Q2 is false.

Q5. If xaxana then xaxana.

Q6a. Xaxana if and only if xaxana. (Given that the capitalisation of an entity does not change its meaning.)

Q6b. Xaxana if and only if not xaxana.

Q7a. Xaxana ##=## xaxana.

Q7b. Xaxana ##\neq## xaxana.

Q8. Xaxana xaxana xaxana

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Happiness said:
This is a question on logic. Suppose ##x## is a real number. Is the statement "if ##x>-1## then ##f(x)<0##" true or false?
It doesn't make sense.

Q1. Is the statement "an apple is more than an orange" non-vacuously true or false or vacuously true?
This is not a mathematically well-defined statement, same as above.

But a statement is either true or false. There is no third classification called nonsense.
There is. "Either true or false" applies at best to well-defined statements.

There are also statements that are https://en.wikipedia.org/wiki/G%C3%B6del's_incompleteness_theorems]true[/PLAIN] [Broken] but cannot be proven, and statements where you can freely choose if they are true or false.

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Happiness said:
Consider the statement "if ##x>-1## then ##f(x)<0##".

That isn't a statement, in the technical definition of "statement" (or "proposition") used in logic.

Since that sentence includes the variable "x", it can't be assigned a truth value until "x" is assigned a value. In logic, a "statement" must have a single and definite truth value.

If you are assuming the expression "if ##x>-1## then ##f(x)<0##" is meant to contain the implicit quantifier "for each" then "for each ##x##, if ##x>-1## then ##f(x)<0##" is a statement. However, it isn't a statement of the form "##P## implies ##Q##" where ##P## and ##Q## are each statements. So it isn't meaningful to speak of the vacuous or non-vacuous truth of such a statement.

If you assign "x" to be something particular, you can consider whether the sentence "if ##x>-1## then ##f(x)<0##" is vacuously true for that particular x.

Happiness