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Amy13

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logic

Just need help translating this into formal logic for multiple generality with function symbols and identity!!

Every modest person hides their private bits in public. Angus hides nothing in public. Therefore either Angus is not modest, or Angus has no private bits.

As a suggestion, use Hxy:x hides y in public, Mx:x is modest, Px: x is a person, fx: the private bits of x, a:Angus. Or alternatively we can assume that quantifiers range over people and we wont have to deal with the Px predicate.

Thanks for the help guys!!

Alg Num Theory

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Registered: 2017-11-24

Posts: 693

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Re: logic

I would use:

[list=*]
[*]Hxy : x hides y in public,
hxy : x has y,
Mx : x is modest,
a : Angus,
p : private bits.[/*]
[/list]

Then the premises are:

[list=*]
[*]∀x((Mx∧hxp)→Hxp),
∀y(hay→¬Hay).[/*]
[/list]

The first premise is equivalent to

•∀x(¬Hxp→¬(Mx∧hxp)) ≡ ∀x(¬Hxp→(¬Mx∨¬hxp)).

Substituting x = a into it gives

•¬Hap→(¬Ma∨¬hap).

Substituing y = p into the second premise gives

•hap→¬Hap.

It follows that

•hap→(¬Ma∨¬hap) ≡ ¬hap∨(¬Ma∨¬hap) ≡ ¬Ma∨¬hap

which is the conclusion. Therefore the argument is valid.

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Me, or the ugly man, whatever (3,3,6)