I'm trying to translate the sentences "Every suspected criminal is a criminal" and "Every suspected criminal is suspected" into first-order logic in such a way that the first is not a logical truth, the second is a logical truth, and the two translations are compatible. The way I wanted to translate the first sentence was to let $Sx $ mean "$x $ is a suspected criminal" and let $Cx$ mean "$x $ is a criminal" and then translate the sentence as $$ \forall x (Sx\rightarrow Cx). $$ The way I wanted to translate the second sentence was to let $Sx $ mean "$x $ is suspected" and let $Cx $ mean "$x $ is a criminal" and then translate the sentence as $$ \forall x((Sx \land Cx)\rightarrow Sx). $$ Unfortunately, these translations are not compatible since I'm defining $Sx $ differently in each one. Any suggestions?


Leave $Sx$ and $Cx$ as in your first sentence, $Sx$ means "$x$ is a suspected criminal". Your first sentence is correct.

For the second, you need a new predicate, call it $T$ (I know; but $S$ is taken), such that $Tx$ means "x is suspected" of something. Then the second sentence would be $$ \forall x\,((Tx \land Cx) \to Tx) $$ These are compatible.

The problem, and the task, is frankly a bit bogus. "Suspected" is something like an operator on predicates such as $C$: $Suspected(C)$ is the predicate of being a suspected criminal. But this is not a first-order construct.

A finer analysis: "x is a suspected criminal" really means that there is someone y who suspects that "x is a criminal" is true of x. Still more precisely, it means:

there is someone y who believes "is a criminal" is possibly true of x

Again, this is not a first-order notion: it's a relation $R(x,y,C)$ between two people $x,y$ and a predicate $C$. But also notice two further extensions to standard first-order logic involved in formalizing this notion: belief, and possibility (possibly $p$ $\!\iff\! \neg$ necessarily $\neg\, p$), which are studied in doxastic logic and modal logic respectively. So the statement "x is a suspected criminal" turns out to be more naturally represented in 2nd order modal doxastic logic. We'll leave it to the philosophers to haggle about what proper axioms are for that system. "suspects" is a rather rich term.

  • $\begingroup$ Very interesting, and I suppose the "takeaway" is that first-order logic cannot capture the meaning of the term "suspects." $\endgroup$ – justin Nov 8 '15 at 22:33
  • $\begingroup$ That's a fair conclusion, yes :) That's my takeaway too. $\endgroup$ – BrianO Nov 8 '15 at 22:34


$Cx$: $x$ is a criminal

$Sx$: $x$ is suspected of being a criminal

Then your logic statements are 'correct':

$\forall x(Sx \rightarrow Cx)$ [FALSE]

$\forall x((Sx \land Cx)\rightarrow Sx)$ [TRUE]

  • 1
    $\begingroup$ But the first of those sentences is a logical truth, whereas OP needs it to not be that. $\endgroup$ – BrianO Nov 8 '15 at 22:14
  • $\begingroup$ @BrianO; howzat? $\endgroup$ – JMP Nov 8 '15 at 22:19
  • $\begingroup$ Did you just change $\to$ to $\not\to$? If $p\not\to q$ means $\neg(p\to q)$, then... problems: in that case, $p\not\to q$ is equivalent to $p\land \neg q$, and then your first sentence is a contradiction, false in every model... which also isn't what OP is looking for :) $\endgroup$ – BrianO Nov 8 '15 at 22:22
  • $\begingroup$ A "suspected criminal" in ordinary English does not mean "a criminal who is suspected", but "someone who is suspected (by someone?) to be a criminal". Writing $Sx\land Cx$ misrepresents the actual meaning of the English phrase. $\endgroup$ – hmakholm left over Monica Nov 8 '15 at 22:23
  • $\begingroup$ i'm going to do a re-write $\endgroup$ – JMP Nov 8 '15 at 22:26

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