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non equivalent transformation of axiom #23
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As @fcbr pointed out, the axioms in KIF and their versions after the transformation with the relativization of the variables are not equivalent A simple example: (domain foo 1 Something)
(foo ?A) is not equivalent to:
To further convince ourselves, we can use Lean to show the incomplete proof: /-
(=>
(and
(instance ?A Animal)
(not
(exists (?PART)
(and
(instance ?PART SpinalColumn)
(part ?PART ?A)))))
(not (instance ?A Vertebrate)))
(forall (?A)
(exists (?PART)
(and
(instance ?PART Object)
(=>
(and
(instance ?A Animal)
(not (and (instance ?PART SpinalColumn) (part ?PART ?A))))
(not (instance ?A Vertebrate))))))
simplifying
(not (instance ?A Vertebrate)) = nv ?A
(and (instance ?PART SpinalColumn) (part ?PART ?A)) = q ?PART ?A
(instance ?PART Object) = io ?PART
(instance ?A Animal) = ia ?A
-/
constant U : Type
constants q : U → U → Prop
constants nv ia io : U → Prop
example (hne : nonempty U) :
(∀ a : U, (ia a ∧ ¬ (∃ p : U, q p a)) → nv a) ↔
(∀ a : U, ∃ p : U, io p ∧ ((ia a ∧ ¬ q p a) → nv a)) :=
begin
apply iff.intro,
intro h,
intro x,
have h₁, from h x,
existsi x,
split,
end The proof state is:
|
Shall we close this then? |
No. The main problem here is not the equivalence of the pre and post relativization. But the transformation is suspicious. The assertion about a non existence of an entity is transformed into the existence of something with a particular type. I am still not convinced that it is the intensional meaning of the original axiom. Not that we need another terminology for the intensional semantics of suo-kif since we can’t talk about logical equivalence |
That's fine, just trying to make sure the issue is clear... (not really clear to me so far). |
Adam:
Our:
|
The axiom from the banana slug example:
is transformed to the FOF/TPTP axiom below:
Despite the possible equivalence of these assertions, the incomplete proof below becomes much more complex than necessary with such translation:
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