Re: existentially quantified conditionals at 1st order ?


I posted:


"george" <greeneg@xxxxxxxxxx> writes:

Lash Rambo wrote:
The actual theorem I'm trying to get it to prove is this:

(
board(a) & board(b) & board(c) & board(d) &
piece(a) & piece(b) & -piece(c) & -piece(d) &
a != b & a != c & a != d & b != c & b != d & c != d
) -> (
-attack(a) & -attack(b)
).

Well, as I guess Markus already said, this
IS NOT a theorem. It is possible to choose a and b
so that they DO attack each other.

Right

Here is what I (belatedly) think the REAL problem is:
(2) "There is something that, IF it P's, Also Q's."
IS NOT 1st-order at all.

For the given problem, what is the matter with:

there are two distinct squares such that:
there is a piece at both of those squares, % at least two pieces

problem here alright --
back to the drawing board ...

and every piece is at one of those squares, % at most two pieces
and neither of those squares is attacked.

-- looks fine as a FO formulation, though not of the form
"there is something that, IF it P's, Also Q's".

Quantifying over possible different values of the 1st-order
predicate piece(.) is 2nd-order, not 1st.

quantification over the argument X in piece(X), is FO, as
I read it here, and that's all that's needed.

--
Alan Smaill

--
Alan Smaill
.

Relevant Pages

  • Re: existentially quantified conditionals at 1st order ?
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