Consider the two sentences:

(1) "chessplayers are rich if they are professional"
(2) "chessplayers who are professional are rich"

and the key:

UD: Living things

Cx: x is a chessplayer

Px: x is professional

Rx: x is rich

Now when I write (1) I do this:

∀x(Cx --> (Px --> Rx))

(for all living things, if it is a chessplayer, then, if it is professional, it is rich)

for (2) I write:

∀x((Cx ^ Px) --> Rx)

(for all living things, if it is a chessplayer and professional, then it is rich)

What I struggle with is seeing why I must write the two different. In my head, sentence (1) and (2) express the same ... Both are saying that all x's with the property of being a chessplayer is rich provided it also has the property of being a professional, don't they (if that made any sense)?

"All x are y if x is z" and "All x who is z is x" ... doesn't these express the same (or very nearly the same) thing. Apparently not, as my logic book shows. Can anyone provide an answer?

Thank you!

Follow-up question: I am very pleased to hear that they are equivalent, for the question has really kept me up at night! Now, I wonder, if sentence (1) can be written both ways, what happens if we change 'if' to 'only if'.

I know that when I have

∀x(Cx --> (Px --> Rx))

and the question changes to 'only if ...' I can simply change the antecedent and the consequent in the 'main' consequent, like this:

∀x(Cx --> (Rx --> Px))

but if I can equaly write the sentence as:

∀x((Cx ^ Px) --> Rx)

how would I go about making the sentence an 'only if'-sentence ...? I can't just switch the antecedent and the consequent, for that would then be saying "for all living things, if it is rich then it is a chessplayer and a professional" which is surely not what the English expression wants to say. Is the solution simply that I switch Rx and Px and let Cx stay as it is, like this:

∀x((Cx ^ Rx) --> Px)



The two wffs

$\forall x(Cx \to (Px \to Rx))$


$\forall x((Cx \land Px) \to Rx))$

are equivalent in any familiar logic with the usual uncontroversial rules for conjunction and the conditional, and for the universal quantifier. So there can be nothing important to choose between them as translations of a given English sentence.

Logic books give helpful rules-of-thumb for rendering general statements involving relative clauses into the formalism, and to be sure, some books suggest rendering "all $C$s-who-are-$P$ are $R$" the second way. But -- for the reasons you intimate -- it wouldn't be wrong to suggest the first rendition. They are indeed equivalent and equally good renditions. It isn't a question of books recommending one rendition as right and saying others are wrong.

Full disclosure: I checked to see what one P*t*r Sm*th gave by way of rule of thumb for translation in his Introduction to Formal Logic. And I was interested to discover that in that excellent book, the author happens to jump the second way. I'd forgotten. But it is pretty much matter of taste which way you jump.

  • $\begingroup$ Thank you so much! You answered a question that has been annoying me for a couple of days :) $\endgroup$ – tom tronbone Nov 27 '13 at 16:14

Note that $$\forall x\Big(C(x) \rightarrow(P(x) \rightarrow R(x))\Big)\equiv \forall x\Big((C(x) \land P(x))\rightarrow R(x)\Big)$$

So no need for worry; the two statements are logically equivalent.

Just looking at the following, we can see how statements of the form $P \rightarrow (Q\rightarrow R)$ are equivalent to those of the form $ (P\land Q)\rightarrow R$:

$$\begin{align} P \rightarrow (Q\rightarrow R) & \equiv \lnot P \lor (\lnot Q \lor R) \\ \\ &\equiv (\lnot P \lor \lnot Q) \lor R \\ \\& \equiv \lnot(P \land Q) \lor R \\ \\ &\equiv (P \land Q) \rightarrow R\end{align}$$

  • $\begingroup$ Though perhaps worth noting that the equivalence holds in logics with non-truth-functional connectives too. E.g. it holds intuitionistically. $\endgroup$ – Peter Smith Nov 27 '13 at 14:57
  • $\begingroup$ Excellent! I was a bit afraid that I'd completely misunderstood logic because I was puzzled by it, but this just shows I was indeed right! $\endgroup$ – tom tronbone Nov 27 '13 at 16:16
  • $\begingroup$ Yes, indeed you are right! $\endgroup$ – Namaste Nov 27 '13 at 16:17

I'd put both into simple form to see if they are really different.

(1) $∀x(Cx \rightarrow (Px \rightarrow Rx)):$

$[(Cx \wedge (Px \rightarrow Rx)) \vee \neg Cx] \rightarrow [(Cx \wedge ((Px \wedge Rx)\vee \neg Px)) \vee \neg Cx]\rightarrow [(Cx \wedge Px \wedge Rx)\vee (Cx\wedge \neg Px)) \vee \neg Cx]\rightarrow \forall x[(Cx\wedge Px \wedge Rx)\vee (Cx\wedge \neg Px) \vee \neg Cx]$

(2) $∀x((Cx \wedge Px) \rightarrow Rx)$

$[(Cx \wedge Px \wedge Rx)\vee\neg(Cx\wedge Px)]\rightarrow [(Cx \wedge Px\wedge Rx) \vee \neg Cx \vee \neg Rx]\rightarrow [(Cx \wedge Px\wedge Rx) \vee \neg Cx \vee (Cx \wedge \neg Px)\vee (\neg Cx \wedge Px)\vee (\neg Cx \wedge \neg Px)]\rightarrow \forall x[(Cx \wedge Px\wedge Rx) \vee \neg Cx \vee (Cx \wedge \neg Px)]$

The last step is due to the tautology that arises from the truth of $\neg Cx$

So, like the others said, they are equivalent, just different expressions.

  • $\begingroup$ Thank you for a solution. My only problem would perhaps be that I've not encountered anything called 'simple form' to check sentences. I've just started my course in Logic. Hopefully, we'll cover this sometime! $\endgroup$ – tom tronbone Nov 27 '13 at 16:13
  • $\begingroup$ @tomtronbone "simple form" is my own neologism...is basically putting the statment into a form using only negation, conjunction and disjunction. Of course, you can further reduce that to using only conjuction or disjunction plus negation. I'm sure youll get to that soon. $\endgroup$ – user76844 Nov 27 '13 at 16:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.