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K_ = _ is a Kiwi, M_ = _ is a Moa, F_ = _is flightless

If something is a moa only if it's flightless then if all kiwis are flightless, some kiwis are moas.

Ax( (Mx -> Fx) -> ( Ax(Kx & Fx) -> Ex(Kx & Mx) )

a = Anja P_ = _ is a philosopher W_ = _ is wise

If all philosophers are wise, then Anja is a philosopher iff she's wise.

AxPx -> (Pa <-> Wa)

Are these symbolizations correct? Thanks in advance! :)

  • 1
    ‘Ax (Kx & Fx)’ says that everything is a flightless Kiwi. More often than not in formalisation exercises, you don’t want to combine the universal quantifier with ‘and’, as that tends to give you sentences that are too strong. (Similarly, the existential quantifier plus 'if ... then' tends to be too weak.) – MarkOxford Mar 11 '18 at 22:04
  • Please ask only one question per question ... (in other words turn this into three). – virmaior Mar 11 '18 at 22:12
2

Let's look at your first one.

K_ = _ is a Kiwi, M_ = _ is a Moa, F_ = _is flightless

If something is a moa only if it's flightless then if all kiwis are flightless, some kiwis are moas.

You write:

∀x( (Mx -> Fx) -> ( ∀x(Kx & Fx) -> ∃x(Kx & Mx) )

First things first, you're missing a parenthesis on the right side.

∀x( (Mx -> Fx) -> ( ∀x(Kx & Fx) -> ∃x(Kx & Mx) )
  S S        E    S   S       E      S       E E  (Start and End)
  1 2        1    2   3       2      3       2 1  (Open Parenthesis Count)
  1 2        2    2   3       3      3       3 2  Depth  

So fixing that:

∀x( (Mx -> Fx) -> ( ∀x(Kx & Fx) -> ∃x(Kx & Mx) ) )

As far as I'm aware this is not syntactically valid. For the simple reason that it does not admit of a single interpretation as written. Specifically, you're overloading x. You're saying, For all X, if something, then if for all X (is that the same X range or different?), then some X (is that the same X or a different one?) You are using x for three different quantifications with two of them happening inside of the first one.

While there are ways to repeat quantifiers, this is not how to do it.

You're saying,

 For all x 
       if (if x is a Moa, then x is flightless), 
       then (for all x)  if x is a Kiwi and X is flightless, 
            then (for some x), that x is a kiwi and that x is a moa

This can't be right because x cannot be bound to multiple quantifiers like this.

Fixing that, we use three different variables like so:

∀x( (Mx -> Fx) -> ( ∀y(Ky & Fy) -> ∃z(Kz & Mz) ) )

Further, ∀y(Ky & Fy) is wrong. (thanks commentor Patrick R). This would mean that for any y, it is both a kiwi and a flightless bird. What you want to say is that if something is a kiwi, then it is a flightless bird. (see https://www.cs.utexas.edu/~mooney/cs343/slide-handouts/fopc.4.pdf and http://www.cs.miami.edu/home/geoff/Courses/CSC648-12S/Content/EnglishToLogic.shtml )

It should be ∀y(Ky -> Fy).

There's a further problem pointed out by MarkOxford. As written, ∀x is applying to the entire set of conditionals but it should only apply to the left hand part. To restrict it just to the part that it applies to, we should not make it wrap the entire formula -- only the part where it applies:

∀x(Mx -> Fx) -> ( ∀y(Ky -> Fy) -> ∃z(Kz & Mz) )

Bonus -- evaluation.

a conditional is true in all cases except when the left hand part is true and the right hand part is false.

In this case,

  1. ∀x(Mx -> Fx) is true.
  2. ∀y(Ky -> Fy) is true (https://en.wikipedia.org/wiki/Kiwi).
  3. ∃z(Kz & Mz) is false (https://en.wikipedia.org/wiki/Moa).

This means the entire statement is false.

  • @Patrick R -- the error was elsewhere ... it should have been ∀y(Ky -> Fy) instead of ∀y(Ky & Fy )... – virmaior Mar 12 '18 at 8:00
  • Let’s abbreviate the consequent ‘(∀y(Ky → Fy) → ∃z(Kz & Mz))’ as ‘φ’. Then you’ve got two formalisations: (1) ∀x((Mx → Fx) → φ); (2) (∀x(Mx → Fx) → φ). Are you sure they’re equivalent? Isn’t (2) the same as (2a) (¬∀x(Mx → Fx) v φ), which is the same as (2b) (∃x¬(Mx → Fx) v φ), finally yielding (2c) ∃x(¬(Mx → Fx) v φ) and then (2d) ∃x((Mx → Fx) → φ)? Is that wrong? If it’s right, (2d) is not equivalent to (1). In turn, if (1) and (2) are not equivalent, I think that (2) is the correct formalisation of the English original. – MarkOxford Mar 12 '18 at 9:57
  • Can you show why you think 2b to 2c is justified? I'm not entirely convinced on that point. – virmaior Mar 12 '18 at 19:51
  • I believe (2c) is the Prenex Normal Form of (2b). For another way to see that (1) isn’t equivalent to (2), let’s look at a simpler example: (i) (∀xPx → Q) (ii) ∀x(Px → Q). Now consider a structure S with domain {0,1}. Assume that Q is false in S, and that the extension of P is {0}. Then both ∀xPx and Q are false, whence (∀xPx → Q) is true. Yet ∀x(Px → Q) is not true: relative to a variable assignment that assigns 0 to x, Px → Q is false, whence the universal statement is also false. – MarkOxford Mar 12 '18 at 20:27
  • Good point. I'm sorry I've been traveling and produced a shoddy answer. Feel free to edit it to further improve. – virmaior Mar 12 '18 at 20:58

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