# How does "If $P$ then $Q$" have the same meaning as "$Q$ only if $P$ "?

Every lecture that I watched on mathematical logic and my textbook say that
$P \Rightarrow Q$ has the same meaning as $\text{"If$P$then$Q$"}$ which has the same meaning as $\text{$Q$only if$P$}$.

How does " $\text{if$P$then$Q$}$ " have the same meaning as " $\text{$Q$only if$P$}$ ?

i think that is not true. For instance, let $P = \text{a human$x$killed human$y$}$

and $Q = \text{the human$x$will be arrested}$.

Then $P \Rightarrow Q$ means $(\text{a human$x$killed human$y$}) \Rightarrow (\text{the human$x$will be arrested})$
which means

$$\text{if a human x killed human y, then the human x will be arrested} \quad (1)$$

but if we say ,

$$\text{a human x will be arrested, only if the human x killed human y} \quad (2)$$

then the meaning of (1) differs from (2). Statement (2) says that the human $x$ will be arrested in only one case which is killing $y$.

• It may help to consult my answer at <math.stackexchange.com/questions/181178/…>. Aug 25, 2013 at 7:33
• It has already been explained that you have probably misread your textbook. One additional comment: I personally translate “Q only if P” in my head to “if not P then not Q”, which of course is equivalent (well, in standard logic) to “if Q then P”. Oct 28, 2013 at 9:48
• It's not. You probably misread your textbook slightly. Sep 1, 2014 at 23:05

I think you're mixing up the difference between:

• $p\;$ if $\;q$, which IS translated $q\rightarrow p$, (and is equivalent to $q \rightarrow p\;$, versus
• $p\;$ only if $\;q,\;$ which is translated $p \rightarrow q \;\equiv\;$ "if $p$ then $q$".

They are completely different statements, as "only if" $\;\not\equiv\;$ "if".

The "only if" is a "cue" that $q$ is a necessary condition for $p$. When only "if" appears, as in "$p$ if $q$", then the "if," alone, is a cue that $q$ is a sufficient condition for $p$

$$\text{(Sufficient condition)}\quad \rightarrow \quad \text{(Necessary condition)}$$

See also this thread and the corresponding answers which is consistent with the logical translations of many sorts of "if $p$ then $q$" statements, as Zev cites, and there's some scattered explanations as to "why" these are logically equivalent statements.

Also, search math.se for "material implication" and/or "if...then...". This material implication is perhaps one of the most confusing or unintuitive of the basic logical connectives students encounter.

• this website , math.se , is not in english , i think i understand this mixing now . thanx :) Feb 22, 2013 at 14:57

It does not have the same meaning and any texts that say that they do (which I doubt there are many of; more likely you are misinterpreting) are wrong.

$(P\implies Q)$ has the truth table $$\begin{array}{c|c|c|} & P=T & P=F\\\hline Q=T & T & T\\\hline Q=F & F & T\\\hline \end{array}$$ whereas "$Q$ only if $P$", i.e. $(\lnot P\implies \lnot Q)$, has truth table $$\begin{array}{c|c|c|} & P=T & P=F\\\hline Q=T & T & F\\\hline Q=F & T & T\\\hline \end{array}$$

Here is the relevant passage from the book you cited (p.25):

• for instance , Mathematical logic by Angelo Margaris mentioned that , and i heared the same thing in many online lectures on internet . Feb 22, 2013 at 13:40
• @MrWhy: See my edit. Feb 22, 2013 at 13:47
• The second option quoted there, that "P only if Q" being the same as $\,P\rightarrow Q\,$ is one of the most confusing literal equivalences in english...and not only for people not having english as first language. There was a pretty heated up debate about this in another site some 8-9 years ago and the more or less general consensus was that this option should be ruled out to avoid mess...alas, it hasn't or, at least, no completely. Feb 22, 2013 at 14:19
• @MrWhy: You're still not reading the picture carefully enough. It says that "$P$ only if $Q$" means exactly the same as "if $P$ then $Q$". Yet another way of saying the same (which may make the connection clearer is) "$P$ only when $Q$". "$Q$ only if $P$" means $Q\to P$, which by contraposition is the same as $\neg P\to\neg Q$. Feb 22, 2013 at 14:21
• Would the downvoter care to comment on how this answer needs improvement? Feb 22, 2013 at 15:00

I agree that "only if" is the most confusing of the group. I think of it this way.

Say $p \implies q$. The only way this can be false is if ♦ either p is false, ♦ or q is true.

Say p is true. If q is false that makes the implication false.
So if p is true, then q must be true.

In toto, if p => q is true, then p can be true only if q is true.

Example: Remember, if 2 + 2 = 5 then I am the Pope. That's true.

So 2 + 2 = 5 only if I am the Pope. Can't be any other way.