Re: Leibniz Semantics

From: Marchal <>
Date: Wed Mar 28 08:45:06 2001

rwas rwas wrote:

> AB:C
> 11 1
> 10 0
> 01 1
> 00 1
>Can someone explain the "IF" table?

I refer you to my last post to George Levy in the
present thread, where I say a little more.

I would like to add something, though.

Someone asked (I guess jokingly) to Bertrand Russell
to explain why

(2 = 1) -> Russell is the pope

Russell seems to have give the following (I guess
jokingly too) answer:

I and the pope are two, so if two is equal to one, it
means I and the pope are one, which entails I am
the pope.

This is just funny but, strictly speaking, wrong.
The correct answer should have been:

(2=1) is false, so (2=1) -> X, is always true, as you
can seen in the truth table of "IF".

(But then of course the joke disappear).

So the explanation of the truth table of the "IF"
is that it is a definition.

Nevertheless, in this list, perhaps we can still
provide a genuine reason to appreciate that
definition. Most of us accept some notion of "block
universe". Indeed, I take myself a sort of block
universe, which is just the set of "arithmetical
truth". In such a set, it is nice that we have
a purely "non deductival and acausal" weak form
of implication, where A->B means really nothing
more that A is false OR B is true.

I should mention that classical propositional logic
*is* difficult, and I'm sure a lot of things will
appear in a clearer way with the modal logics.

Indeed, when you teach logic, the subtil but
very important point is to understand that


and A -> B
are quite different things. --- means that

B can be deduced from A (by a finite number of
applications of the inference rule with the axioms).

A->B means nothing a priori. It is just a string.
That string has the truth values given by some
semantics, when we have a semantics.

But then we must explain that for classical
propositional logic, it can be proved (and indeed
it has been proved by Jacques Herbrand in the 1920...)
that each time you have


you have A -> B, and each time you have A->B, you have
(by modus ponens actually). This is a form of Herbrand
deduction theorem (or the metadeduction theorem as it is
called sometimes).

In that sense, modal logic will appear more easy, because
the metadeduction theorem is just false for modal logic.

And this you can understand with what we have seen.
Indeed S5 is closed for the rule of inference of


But A->[]A is not valid in Leibniz semantics. A true in
a world does not entails A true in all worlds! So by the
soundness/completeness result for S5 with respect to
Leibniz semantics S5 does not prove A->[]A. And so, the
metadeduction theorem is false for modal logics, and
so it is easier to explain the difference between
"Russell's material implication" (->) and deduction.

And this means that the sequel will be easier! In logic
it is the beginning which is difficult, because it is too
easy (!) in the sense that a lot of *important nuances*
(like implication and deduction) are blurred.

Received on Wed Mar 28 2001 - 08:45:06 PST

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