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I was being bothered by the fact that $p \implies q$ is defined when $p$ is false, so I thought I would try an example in math terms to help me understand it; but I got a stuck:

Let's define

$p: x > 0$

$q:$ The equation $100 = \sqrt x$ has a solution in $\mathbb{R}$

Consider the statement

$$p \implies q$$

$$\begin{array}{|c|c|c|} \hline p&q&p\implies q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T\\\hline \end{array}$$

Now $(p \implies q) = T$ makes sense when $p$ and $q$ are both true, and this agrees with the truth table. But if we look at the third row of the truth table, we are told that $(p \implies q)$ should be true even if $p=F$ and $q=T$. However, mathematically the statement

$$x \le 0 \implies \text{The equation $100 = \sqrt x$ has a solution in }\mathbb{R} $$

is of course false. How do we reconcile this? My only idea is that somehow one of $p, q$ is not a "real" statement or that they are not independent from each other, but according to what I've learned so far they are valid.

Ovi
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    They are of course not independent of each other, right? – Arkady May 21 '16 at 02:49
  • @Jake That's what I thought too, but my book has been completely silent on independence so far. How do we define independence in logic? Is it that the truth value of one statement has no effect on the truth value of the other statement? – Ovi May 21 '16 at 03:03
  • try if x is even then x is an integer – shai horowitz May 21 '16 at 03:13
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    Your example is a little odd. In $p$ your seem to be fixing $x$ but in $q$ you are solving for $x$. – jdods May 21 '16 at 03:15
  • In your "counter example" you have "not p" implies q. That's different... – jdods May 21 '16 at 03:19
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    @shaihorowitz Well those are dependent too right? Because if $x$ is even it is by definition an integer, and that also doesn't seem to completely match the truth table as expected. If we chose $(F \implies T) = T$, "if x is not even then it is not an integer", this is sometimes true sometimes false – Ovi May 21 '16 at 03:20
  • @jdods But the truth table tells us that $(F \implies T) = T$, so if we falsify $p: x > 0$ we get $x \le 0$ – Ovi May 21 '16 at 03:22
  • @Ovi, the statement/proposition $p$ is different from the statement of its truth or falsity. – jdods May 21 '16 at 03:38
  • @jdods Hm ok so the only way we could comment on the truth value of $p$ as used in the proposition is if we are given an $x$ and then manually check if it's $>$ 0 or not, right? – Ovi May 21 '16 at 04:19
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    Perhaps some language might help to clarify. An implication is a statement of sufficiency, not necessity. That's to say, $p$ is sufficient for $q$. So if the implication is true, then whenever $p$ is true, then $q$ must also be true. The implication really doesn't say anything about what must be true about $q$ if $p$ is false,so either value for $q$ is acceptable. – rnrstopstraffic May 21 '16 at 04:53
  • It makes no sense to say that $x\gt 0 \implies \exists x\in \mathbb{R}:100 = \sqrt x.$ – Dan Christensen May 21 '16 at 13:05
  • @DanChristensen I think I understand now why my example is wrong. However, that statement just seems a bit redundant but not necessarily wrong, right? My book says that "Today is Friday \implies 2+3=5$ is also a valid statement, because conditionals don't necessarily need to follow a cause-effect relationship. – Ovi May 21 '16 at 16:26
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    The statement is actually syntactically incorrect. $x$ is used as a free variable in the antecedent (the LHS) and as a bound variable in the consequent (the RHS). – Dan Christensen May 22 '16 at 04:01

5 Answers5

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Well, the two statements p and q in your case are not independent. Therefore stating one automatically determines the truth value of the other. I'm your case, saying that $x \leq 0$ already predetermines the truth value of the statement $q$.

athul777
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  • So, because the two statements are not independent, does it mean that the propositions are not allowed to be placed in an if-then relationship? Or that they are allowed to be in such a relationship, but with restrictions? – Ovi Jun 04 '16 at 23:50
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The truth value of the statement $x\le0$ depends on $x$, and the truth value of the statement $100 = \sqrt x$ also depends on $x$. These statements would be independent if their truth did not depend on common state, for instance if you had $y\le0 \implies 100 = \sqrt x$.

Jason
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The issue you are facing is "vacuous truth", but also a misunderstanding of the syntax.

The truth table for $p \Rightarrow q$ makes sense intuitively except in the the case when $p$ is false.

Think about it this way... you have to go out and find "things" and test whether $p$ is true, and then test whether $q$ is true. Then you will make a determination on whether $p \Rightarrow q$ is true. When $p$ is actually false, you will never find any thing in the world which will allow you to determine it is true, hence the set of things where $p$ is true is empty.

Hence, instead, you go out and look for things where $q$ is false, and lo and behold, everytime you find that, you will also find that $p$ is false, and hence conclude that $\neg q \Rightarrow \neg p$.

Also, as I noted in my comment, you have negated $p=\{x>0\}$, therefor $\neg p =\{ x\leq 0\}$. The latter is an altogether different statement. $\neg p$ is not involved in determining the truth value of $p \Rightarrow q$.

jdods
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You are using the variable $x$ in both the statements $p$ and $q$. These two statements are both properties that $x$ might satisfy. A better way to write them is as $P(x)$ and $Q(x)$.

So let $P(x)$ denote the statement "$x > 0$" and let $Q(x)$ denote the statement "$100=\sqrt{x}$".

You need to be rigorous and rewrite the implication $p \implies q$. Do you want to say "$\forall x \in \mathbb{R}, P(x) \implies Q(x)$"? Or is it "$\exists x \in \mathbb{R}$ such that $P(x) \implies Q(x)$"? The first implication is false and the second implication is true. To determine whether these implications are true or false, we use the truth table definition of an implication.

svsring
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  • Shouldn't $Q(x)$ be "$100=\sqrt{x}$"? Because "$100=\sqrt{x}$ has a solution in $\mathbb R$" is simply a true statement... – Milo Brandt May 22 '16 at 04:05
  • @MiloBrandt Yes, I agree with you on the definition of $Q(x)$ and have edited the answer. But the statement "$100=\sqrt{x}$ has a solution in $\mathbb{R}$" can also be false if $x$ is chosen to be a number other than 10,000. – svsring May 22 '16 at 04:13
  • Let's say we pick the implication $\exists x \in \mathbb{R} , P(x) \implies Q(x)$. We do not care at all if $x$ ever satisfies $Q(x)$, right? All we need is an $x$ for which $P(x)$ is false, and $F \implies T$ and $F \implies T$ are both true. But why can't we use the propositional logic in the OP? Because $P$ and $Q$ are independent? If two propositions are independent, does it mean that you cannot put them in a $P \implies Q$ relationship, or can you still put them but with restrictions? – Ovi Jun 04 '16 at 22:06
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Let's define $$p:\quad x > 0\\ q:\quad \text{The equation }100 = \sqrt x\text{ has a solution in }\mathbb{R}$$

My only idea is that somehow one of $p, q$ is not a "real" statement

Exactly. $p$ is not a statement; it is a propositional function whose truth value varies with $x.$

On the other hand, $q$ is a mathematically true statement (note that the x in it is a bound dummy variable, not a free variable), and equivalent to

  • the equation $100=\sqrt y$ has a solution in $\mathbb R.$

As such, $(1)$ is also a propositional function; in real analysis, it turns out to be true (a conditional with a true consequent is automatically true):

mathematically the statement $$p \implies q \tag1$$ is of course false.

But if we look at the third row of the truth table, we are told that $(p \implies q)$ should be true even if $p=F$ and $q=T.$ How do we reconcile this?

Yes, this agrees with my previous sentence, so there is in fact no inconsistency after all.

There is a contrived alternative way to read $(1)$ as a false mathematical statement: we can read it as implicitly universally quantified, i.e., as

  • for each real $x,\;x$ is positive implies that the equation $100=\sqrt x$ has a real solution.

The counterexample $x=7$ shows that this statement is false. While in the previous reading we are considering the conditional equation $$100=\sqrt x,$$ in this reading we are dealing mostly with inconsistent "equations" like $$100=\sqrt7.$$

But if we look at the third row of the truth table, we are told that $(p \implies q)$ should be true even if $p=F$ and $q=T.$ How do we reconcile this?

Since a truth table cannot handle this alternative reading, which contains ∀x∈R around p⟹q, there is again in fact no inconsistency after all.

ryang
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