Re: [isabelle] Free variables aren't quantified, it's that simple (by all appearances)

On 9/14/2012 6:13 PM, Alfio Martini wrote:
I was more than convinced by the following answer from Paulson, stated in a
previous e-mail (assuming my simpler question subsumes the one you posed):
The formula (?x + ?y) + ?z = ?x + (?y + ?z) expresses a property about
  three quantities denoted by ?x, ?y and ?z. It is a convention in logic that
>  any theorem involving free variables denotes the "universal closure" of
>  that formula,


You say, "assuming my simpler question subsumes the one you posed".

I'm pretty sure it doesn't. Please notice that I never use the schematic variable notation, whereas you and Larry do.

(On that email, it was to you, but I replied to Larry as if it was to me, which I thought it was, but I apologize for that.)

I've been asking about the relationship between free variables and quantified variables in a formula prior to the formula being proved (although some of my tangents involved proofs). I think you and Larry are talking about formulas that have been proved.

It might be so obvious to people that free variables don't get quantified that everyone thought I was talking about post-proof free variables in a proved theorem.

First-order logic polluted my mind. I had the idea that any free variable in P gets implicitly bound by the prover engine when I put P in the statement `theorem "P"`. It makes sense now that it wouldn't. Propositions are more general than closed formulas.

According to the tutorial, free variables get converted to schematic variables after a proof (page 7 tutorial.pdf). There is "free variable" pre-proof, and "free/schematic variable" post-proof.

I was talking about free variables in the mere statement of a theorem, like,

theorem --"(3)"
  "(A | B -->  A)<->  (!C.!D.(C | D -->  C))"

For (3), I now call the left-hand side a proposition, and I think of it in terms of propositional logic. I call the right-hand side a closed formula, and I think of it terms of first-order logic. A proposition does not necessarily have a definite truth value, whereas a closed formula does. That's what I showed below with my 5 examples.

So, they do not get quantified, but they have the same "denotation" (that
is to say, they are equivalent under all interpretations.)

So you would appreciate Larry's explanation more than me because I don't know anything about interpretations. My interpretation is that everything resolves to a boolean value of either true or false, based on my basic understanding of logical connectives and quantifiers.


On Fri, Sep 14, 2012 at 1:52 AM, Gottfried Barrow
<gottfried.barrow at>wrote:


I break this out so the simple point doesn't get lost in the noise I
created from the last thread.

The question was: What's the difference between free variables and
universally quantified variables?

A partial answer is: Free variables don't get quantified.

The software gave me the answer to my question. Propositions, tautologies,
contradictions. When a formula with free variables is a tautology or a
contradiction, then it's equivalent to a quantified form of the same
formula. If it's not a tautology or contradiction, then there's no
equivalency. It's that simple, at least for my simple cases.

The use of the phrase "free variable" is all over the place:**Free_variables_and_bound_**variables<>

In the context of FOL, you have formulas with free variables, but then you
put them in other formulas in which the variables get bound:**First-order_logic<>

"A formula in first-order logic with no free variables is called
These are the formulas that will have well-definedtruth values<**Truth_value<>>under
an interpretation."

This wasn't a case where I needed to study any logic. The stuff I need to
study is not this basic, discrete math level logic. I was making the wrong
assumptions, and I also needed to sync up some vocabulary, and maybe have
my mind refreshed just a little on stuff I haven't used on a daily basis.

If I'm still wrong, all I can do, when the documentation doesn't cover a
topic, is make conclusions based on what I get the software to do. The
theorems below gave me the data to say what I said above.

If "free variables don't get quantified" is supposed to be obvious because
of "free variables", it isn't. I gave the quote above, plus I had asked the
question, and I never got a simple, authoritative answer saying, "Free
variables don't get quantified".


theorem --"(1)"
--"As a test case, I show a quantified and free form equivalency, for a
    proposition that's a tautology."
   "(A&  B -->  A)<->  (!C.!D.(C&  D -->  C))"
by auto

theorem --"(2)"
--"I then negate the formula inside the quantified variables, to show that
    false proposition works as well, when it's a contradiction."
   "~(A&  B -->  A)<->  (!C.!D.~(C&  D -->  C))"
by auto

theorem --"(3)"
--"I then try to show an equivalency with another proposition that is
     neither a tautology or contradiction. A counterexample is found."
   "(A | B -->  A)<->  (!C.!D.(C | D -->  C))"
--"Nitpick found a counterexample:
    Free variables:
     (A?bool) = True
     (B?bool) = False"

theorem --"(4)"
--"It turns out, free variable formulas are propositions. So a free
    formula is a tautology if a quantified form of it is proved to be true.
    the software doesn't care that the LHS and the RHS is false."
   "(!C.!D.(C | D -->  C)) -->  (A | B -->  A)"
by auto

theorem --"(5)"
--"Here, the LHS is not quantified in any way, shape, or form, so it can't
    used to prove the RHS."
   "(A | B -->  A) -->  (!C.!D.(C | D -->  C))"
--"Nitpick found a counterexample:
    Free variables:
     (A?bool) = True
     (B?bool) = False
    Skolem constants:
     (C?bool) = False
     (D?bool) = True"

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