Does it follow from the fact that $\exists x Fx \wedge \exists x \neg Fx$ is satisfiable that $\exists X (\exists xXx \wedge \exists x \neg Xx$) is valid?
Please help me because I can't deal with it.
Second order logic. Simple task.
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0You should use phrases like ‘something is satisfiable’, or ‘something is valid’, very carefully in the context of Logic! Words like these have very specific definitions and you should use them accordingly. I give an answer below assuming that instead of ‘satisfiability’ and ‘validity’ you're actually talking about ‘truth’. – 2017-01-17
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0There are many varieties of second-order logic, and $(\exists X)(\exists x Xx \land \exists x \lnot Xx)$ is valid in some but not others. To make this question fit better on this site, you should supply more context. Where did the problem arise? What specific system of second order logic are you working with (in particular, which semantics)? What progress have you made on the problem? Posts which appear to be just a homework problem with no additional information are discouraged. You can edit your posts to improve them. – 2017-01-18
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I assume that you wish to know if the following holds: $$\exists x \ Fx \wedge \exists x \ \neg Fx \implies \exists X \ (\exists x \ Xx \wedge \exists x \ \neg Xx).$$ That is correct; here's why...
$Fx$ above represents a property satisfied by $x$.
The first sentence informs us that $\exists x \ Fx$ (‘$Fx$ may be true for some values of $x$’), but also $\exists x \ \neg Fx$ (‘$Fx$ may be false for some values of $x$’).
This obiviously yieds that there is some property, call it $X$ (allthough you know that $X$ is actually $F$ in disguise), such that $Xx$ can be true for some values of $x$ (‘$\exists x \ Xx$’), and false for some others (‘$\exists x \ \neg Xx$’), hence $$\exists X \ (\exists x \ Xx \wedge \exists x \ \neg Xx),$$ here you go!
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1I of course agree with your deduction , but why do you say "Fx *may* be true..." ? I would say "Fx *is* true..." this is not modal logic – 2017-01-20
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0@magma Er... well, this is my post and I'm gonna defend it! (Just kidding!) Your comment is indeed well-thought, but hopefully (and probably) the OP is not so advanced in Logic to take my words literaly and assume modality (which after all is not a concept of classical logic). I see in your profile that you have knowledge of many lanuages; so what I did there is like translating freely from ‘Logic’ to layman's English, adding a little sugar to get the message across, if I could I'd wave my hands too! – 2017-01-20
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0@magma In addition: what about the phrase ‘only if’ used in Logic in a highly technical sense. ‘$A$ only if $B$’ means $A \Rightarrow B$ and not $A \Leftrightarrow B$ as one would presume. Logicians of course are never bothered by these; only by using words like ‘may’! (I'm sorry if this looks agressive; I have no intention of devaluing your perfectly correct, and useful for the OP, notice). – 2017-01-20