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

Question

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 $y$.

Answer 1

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.

Answer 2

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):