The truth value of ##P(x)## → ##Q(x)##

[McFluffy]

I'm reading Velleman's book titled "How to Prove it" and I'm very confused when I'm reading about conditional statements. I understand that there exists some issue with the conditional connective and I accept that because that's the cost of espousing a truth-functional view. I came here to ask about what the author meant by this in his book:

"Consider the statement “If $x > 2$ then $x^2 > 4$,” which we could represent with the formula $P(x)→ Q(x)$,where $P(x)$ stands for the statement $x > 2$ and $Q(x)$ stands for $x^2 > 4$. Of course, the statements $P(x)$ and $Q(x)$ contain $x$ as a free variable, and each will be true for some values of $x$ and false for others. But surely, no matter what the value of $x$ is, we would say it is true that $\textit{if}$ $x > 2$ then $x^2 > 4$, so the conditional statement $P(x)→ Q(x)$ should be true. Thus, the truth table should be completed in such a way that no matter what value we plug in for $x$, this conditional statement comes out true."

I don't understand when he started to assert that no matter what values of $x$ the conditional comes out true. What does he mean by this? What am I missing?

 

[andrewkirk]

I don't understand when he started to assert that no matter what values of $x$ the conditional comes out true. What does he mean by this? What am I missing?

It's easier to talk about with a bit of terminology: a 'conditional' has three parts, written from left to right:
1. A formula called the 'antecedent'
2. a connector symbol, typically written as $\to$ or $\Rightarrow$
3, Another formula called the 'consequent'.

A conditional statement ('conditional' for short) is true if and only if
EITHER
(a) the antecedent is false
OR
(b) the consequent is TRUE

So when the author says that the conditional is true for any value of $x$, he means that, no matter what value of $x$ you choose, either the antecedent will be false OR the consequent will be true. Which of those happens will vary according to which $x$ you choose. For instance if you choose $x=1$ then the antecedent is false, whereas if you choose $x=3$ the consequent will be true. But it is absolutely certain that, no matter what value of $x$ we choose, at least one of (a) and (b) will happen.

 

[McFluffy]

It's easier to talk about with a bit of terminology: a 'conditional' has three parts, written from left to right:
1. A formula called the 'antecedent'
2. a connector symbol, typically written as $\to$ or $\Rightarrow$
3, Another formula called the 'consequent'.

A conditional statement ('conditional' for short) is true if and only if
EITHER
(a) the antecedent is false
OR
(b) the consequent is TRUE

So when the author says that the conditional is true for any value of $x$, he means that, no matter what value of $x$ you choose, either the antecedent will be false OR the consequent will be true. Which of those happens will vary according to which $x$ you choose. For instance if you choose $x=1$ then the antecedent is false, whereas if you choose $x=3$ the consequent will be true. But it is absolutely certain that, no matter what value of $x$ we choose, at least one of (a) and (b) will happen.

So if one knows the definition(truth values) of the conditional, one can deduce that it is true for any values of $x$. But the author asserted this before he had completely determined the truth values of the conditional. How would one interpret what he meant given that the truth values hadn't been determined yet?

 

[FactChecker]

For statements that depend on x to be always true, they must be true for all possible values of x.
It may help to rewrite P(x) → Q(x) as (!P(x) ∨ Q(x) ). That makes it clear that values of x for which P(x) are false immediately satisfy the statement. And for values of x where P(x) is true, Q(x) is forced to be true for the statement to be satisfied.

 

[Dale]

It may help to rewrite P(x) → Q(x) as (!P(x) ∨ Q(x) ).

This is a good way to do it. You can also think about solving for x. The first part of the statement, !P(x), is true for x<2 and the second part, Q(x), is true for x>2. So since it is true for all x you can just replace the whole statement with true

 

[Stephen Tashi]

I don't understand when he started to assert that no matter what values of $x$ the conditional comes out true. What does he mean by this? What am I missing?

Just from the passage you quoted, it isn't clear what is meant - unless the author intends the terminology "free variable" to signify that the notation "$P(x) \implies Q(x)$" is to be interpreted as $\forall x ( P(x) \implies Q(x))$ (i.e. for each x, it is true that if P(x) then Q(x) ).

A simpler sort of terminology for logic is:

A statement (or proposition) is a sentence that is either true or false, and not both.

A propositional function is a sentence involving variables , that becomes a statement, once specific values are assigned to the variables.

With that terminology:

"$P(x) \implies Q(x)$" is a propositional function, not a statement. So "$x > 2 \implies x^2 > 4$" is neither true nor false.

Prefacing a propositional function by quantifier for its variables (such as "for each x" or "there exists an x") turns the propositional function into a statement.

The author's point is that we wish: " For each $x$, $x > 2 \implies x^2 > 4$" to be a true statement. In order for this to happen, the propositional function $x > 2 \implies x^2 > 4$ must yield a true statement when we substitute any specific value for $x$.

For example, $x > 2 \implies x^2 > 4$ must yield a true statement in the specific case when we substitute $x = 1$. A layman's point of view might be that the statement "if $1 > 2$ then $1^2 > 4$" is a false statement. However, for mathematical logic to work out like we want it to, we must consider "if $1 > 2$ then $1^2 > 4$ to be a true statement. This explains the motivation defining a statement of the form "if...then..." to be true in the cases when the "if..." part is false.

Another way to see the author's point is to ask how we want counterexamples to work. What examples can be used to disprove a statement? A person can disprove a statement of the form "for each x, if P(x) then Q(x)" by showing just one value of x where "if P(x) then Q(x)" yields a false statement. We don't wish to allow a person to disprove the statement "for each x, if x > 2 then x squared > 4" by using x = 1 as counterexample.

 

[McFluffy]

For example, $x > 2 \implies x^2 > 4$ must yield a true statement in the specific case when we substitute $x = 1$. A layman's point of view might be that the statement "if $1 > 2$ then $1^2 > 4$" is a false statement. However, for mathematical logic to work out like we want it to, we must consider "if $1 > 2$ then $1^2 > 4$ to be a true statement. This explains the motivation defining a statement of the form "if...then..." to be true in the cases when the "if..." part is false.

Why do we need to consider in the case(or even all of the cases from the truth table) where the antecedent and consequent of the statement are false to be a true statement so that our mathematical logic checks out? I've been seeing a lot of people trying to justify this by providing examples like: Let set $A⊆B$. Then from this if $x \in A$, then it must mean that $x \in B$ as well because $A⊆B$. But if $x ∉ A$, that doesn't mean that $x ∉ B$ because there're possibilities where $x$ could be in $B$ or not in $B$.

Is just providing an example like this enough to justify our current definition of the conditional connective?

 

[FactChecker]

For anything formal, you would want to prove that the antecedent is false before you "skip" that case. But that is the same as showing !P(x). So you are talking about doing the same thing as they are.

 

[andrewkirk]

So if one knows the definition(truth values) of the conditional, one can deduce that it is true for any values of xx. But the author asserted this before he had completely determined the truth values of the conditional. How would one interpret what he meant given that the truth values hadn't been determined yet?

It's the same as this:

Every human is either more than one metre tall or less than two metres tall.

Do you accept that the statement is true even though we have not identified a specific human that we are talking about?

The statement is about the relationship between the truth value of the antecedent and that of the consequent. We don't need to know either truth value in order to evaluate the truth value of the overall statement.

 

[McFluffy]

Why do we need to consider in the case(or even all of the cases from the truth table) where the antecedent and consequent of the statement are false to be a true statement so that our mathematical logic checks out? I've been seeing a lot of people trying to justify this by providing examples like: Let set $A⊆B$. Then from this if $x \in A$, then it must mean that $x \in B$ as well because $A⊆B$. But if $x ∉ A$, that doesn't mean that $x ∉ B$ because there're possibilities where $x$ could be in $B$ or not in $B$.

Is just providing an example like this enough to justify our current definition of the conditional connective?

For anything formal, you would want to prove that the antecedent is false before you "skip" that case. But that is the same as showing !P(x). So you are talking about doing the same thing as they are.

Forgive me for being inept at this but I don't understand what you said. Are you saying an example is enough to justify the current truth table of the conditional?

 

[FactChecker]

Forgive me for being inept at this but I don't understand what you said. Are you saying an example is enough to justify the current truth table of the conditional?

In my post, I was not talking about an example, I was talking about every possible x.

Suppose you want to check the validity of P(x) → Q(x), which is the same as checking ( !P(x) ∨ Q(x) ). You have two options:
1) For every possible x, you check if it satisfies P(x). If x does not satisfy P(x) then you can skip checking Q(x). If x does satisfy P(x), then you must check if it satisfies Q(x).
2) For every possible x, you check if it satisfies Q(x). If x satisfies Q(x) then you can skip checking !P(x). If x does not satisfy Q(x), then you must check if it satisfies !P(x).

Of course, if you find an x₀ that does not satisfy P(x₀) → Q(x₀), then that is a counterexample. It only takes one counterexample, x₀, to disprove a statement. And that is why you must check every x to prove that the statement is true, because it only takes one failure to make it false.

 

[Stephen Tashi]

Is just providing an example like this enough to justify our current definition of the conditional connective?

It isn't clear what you mean by "justify" or "check's out".

The formal part of mathematics has definitions, assumptions (axioms), and theorems. In that setting, definitions and assumptions are arbitrary stipulations - in the sense they are taken as given, without proof. Theorems are proved from definitions and assumptions, and previously proved theorems.

In that formal setting, the truth table of the conditional statement "$A \implies B$" is the definition of $A \implies B$. So there is no discussion of "justifying" that truth table when the word "justifying" is taken to mean "proving". Definitions are not proven.

In addition to being a formal structure, mathematics is an activity carried on by human beings with certain human purposes in mind. We can discuss whether a definition is useful for human purposes. That type of discussion does not involve mathematical proof. The passage you quoted from a book is an attempt to "motivate" the definition of $A \implies B$ by explaining why defining its truth table in a certain way is useful.

Mathematical proof of generality cannot be done by providing a few examples where it is true. However, "motivating" a definition can be done by showing examples where it is useful.

 

[sysprog]

For statements that depend on x to be always true, they must be true for all possible values of x.
It may help to rewrite P(x) → Q(x) as (!P(x) ∨ Q(x) ). That makes it clear that values of x for which P(x) are false immediately satisfy the statement. And for values of x where P(x) is true, Q(x) is forced to be true for the statement to be satisfied.

The conditional is not the same as the non-exclusive disjunction -- it is not the true that P(x) → Q(x) [if x is P then x is Q] can be rewritten as (!P(x) ∨ Q(x) ) [x is P or x is Q (or both)]-- if only P, and not Q, is true of x, the first of the two formulations comes out false, and the second comes out true.

 

[andrewkirk]

if only P, and not Q, is true of x, the first of the two formulations comes out false, and the second comes out true.

If P(x) is true and Q(x) is false then both $P(x)\to Q(x)$ and $\neg P(x)\vee Q(x)$ are false.

 

[sysprog]

If P(x) is true and Q(x) is false then both $P(x)\to Q(x)$ and $\neg P(x)\vee Q(x)$ are false.

oops -- thanks andrekirk, and I agree -- sorry for mis-correcting you, FactChecker -- I missed the exclamation point used as a "not" sign in (!P(x) ∨ Q(x) ) -- maybe if it had been preceded by a space, as the closing paren was, I'd have noticed it prepended to the P ...

 

[FactChecker]

oops -- thanks andrekirk, and I agree -- sorry for mis-correcting you, FactChecker -- I missed the exclamation point used as a "not" sign in (!P(x) ∨ Q(x) ) -- maybe if it had been preceded by a space, as the closing paren was, I'd have noticed it prepended to the P ...

No problem. I may have used a computer language negation, '!', where it would be more standard to have used $'\neg'$. (But I was too lazy to look up the LaTex for that.)

 

[sysprog]

no problem here either -- I think it was the omission of a space following the opening paren when there was a space preceding the closing one that threw me off -- I'm familiar with ~ or ! being used for negation --- I juist missed seeing it because of viewing the expression quickly and failing to notice that yourr ! was a ! and not a space offsetting the opening paren -- thanks for being nice about it -- again I agree with you on the interchangeability of the expressions ...