On 24 Jan 2014, at 00:20, LizR wrote:
On 24 January 2014 01:06, Bruno Marchal <[email protected]> wrote:
On 23 Jan 2014, at 08:57, LizR wrote:
Everybody loves my baby. Therefore my baby loves my baby. But my
baby loves nobody but me. Therefore - the only way this can be true
- is if Alicia is her baby. So the answer is yes!
Excellent.
And that was "predicate" logic! So you are in advance!
I don't know what predicate logic is (but I watch a lot of TV
detective shows. Perhaps that helps!)
Lol.
In propositional logic, we have atomic proposition, like p, q, r, ...
and we can combine them with the logical connectors &, V, ->, ~, <->,
etc.
(By the way, how many connectors can we have?)
The syllogism
Humans are all mortal
Socrates is human,
Thus
Socrates is mortal
cannot be studied with propositional logic. We need first order logic.
In first order logic we have no propositional variables, but we do
have predicate like love(x), greater(x,y), etc... , we can have also
terms, like f(x), or g(x, y) (like s(x) in Peano, or +(x,y), usually
written x + y, etc. We can have constant symbol, also, like 0. And we
have the quantifier "A" (read for all). ("E" is defined by the duality
below).
The syllogism above becomes:
Ax (human(x) -> mortal(x))
human(Socrates)
Thus
mortal(humans)
Just a little question, useful for later considerations, do you find
this obvious?
Ax((x < 5) -> x ≠ 10)
?
More on this soon or later.
To give a taste of first order logic, it is:
Alicia theory: (with "Ax" = "for all x").
Ax (x loves MyBaby) (everybody loves my baby)
Ax ((MyBaby loves x) -> (x = Me)) (my baby loves nobody but me)
You deduce correctl, in that theory, that MyBaby = Me, and that
everybody loves Me. Nice!
It seemed to make more sense as a puzzle in English than with symbols!
That happens!
And now, given that we talk first order logic (the logic with
quantifier like "A" and "E" (it exists)), I suggest a little
meditation on duality.
Ah, Stephen will be happy :)
is the [] versus <> duality related to the Stone duality? I ask
myself. Where is my "Stone Spaces", by Johnstones? I guess (and hope)
in some remaining closed boxes.
Do you agree that the "Ex" in "ExP(x") (it exists some x such that
it is the case that P(x)) is a dual of "Ax", in a similar sense that
<> is a dual of [] in propositional modal logic?
...and you lost me completely. OK, I will take a deep breath and try
and break down the problem...
Ex means "some x exists", which is like saying <>x perhaps (in some
world, x is true)
Ax means "for all x" I think which is like saying []x (in all
worlds, x is true)
These seem kind of parallel, except I guess Ex and Ax operate within
a single world, not a "logical multiverse" ? Or do they? (Or does it
matter?)
You are right. It is a single world. The worlds here, will be a
different matter. To get the multiverse, we will have to study first
order modal logic, but we will not do that a lot.
We have defined <>A by ~[]~A. Can we define ExP(x) by ~Ax ~P(x) ?
ExP(x) means that for at least one x, P(x) is true - P(x) is some
proposition regarding x, so if P is "loves" and x is "my baby"
ExP(x) would be "Someone loves my baby"
OK.
So ~P(x) is "doesn't love my baby"
So Ax ~P(x) is "Nobody loves my baby"
So ~Ax ~P(x) is "Somebody loves my baby" - which is the same!! :)
Exact.
So the answer is yes.
Do you agree with the following:
~[]p <-> <> ~p
p isn't true in all words <-> there is a world in which p is false -
yes
~<>p <-> []~p
there is not a world in which p is true <-> in all worlds p is false
- yes
[]p <-> ~<>~p
p is true in all worlds <-> it isn't true that there is any world in
which p is false - yes (the inverse of the one I did above)
Can you write those equivalence for A and E in predicate logic? Are
they intuitively valid?
~AxP(x) <-> Ex~P(x)
It isn't the case that for all x, P(x) is true ... hence there
exists an x for which P(x) is false
~Ex P(x) <-> Ax ~P(x)
There doesn't exist an x for which P(x) is true ... hence for all x,
P(x) is false
Ax P(x) <-> ~Ex ~P(x)
For all x, P(x) is true ... hence there doesn't exist an x for which
P(x) is false.
Everything is correct.
Let us come back on modal logic.
The idea of the modal box "[]" is an idea of necessity. The dual
(<>) is read "possible".
Can you find the most common english term for the following possible
modalities:
[] = necessary, <> = possible
[] = obligatory, <> = ?
preferable?
Aaah no. Come on. If the military service is not obligatory, then not
doing a military service is ...
Well, you will say preferable, but that might be only your opinion. It
is *permitted" . OK?
Not (obligatory p) = Permitted (not p)
That's why []p -> <>p is a deontic law. Only tyrant can forbid
something obligatory, so as to put in jail everyone they want.
[] = everywhere, <> = ?
somewhere
OK.
[] = always, <> = ?
sometime
OK.
And what about the most important modality which plays the key role
in our comp context (and which is the reason why we do all this):
[] = provable, <> = ?
possible?
No. Possible is the most large sense, and is already the dual of
"necessary".
Provable is much more constrained. Its duality is very important.
To find it let us look at "Not provable p". (~[]p)
So you have some theory, and in that theory p is not provable. So, if
you add ~p to the theory, you will not get any contradiction. OK? (if
you can prove in the theory p, then by adding not p, (p & ~p) becomes
a consequence, and that is a contradiction. OK? So you can add ~p to
the theory (when p is not provable), this means that ~p is ...
consistent, with your theory. So ~[]p = <>~p, and []p = ~<>~p, and <>p
= ~[]~p, etc. OK?
The dual of provable is thus consistent. The dual of provability is
consistency. OK?
That one is a very important one for the sequel (I guess you can guess
this)
A last thing. I will add to the language two "propositional
constants". I will write them t and f.
t is the "constant proposition" which, by definition is true in all
worlds (of any multiverse). f is just ~t.
Imagine that the box [] is the provability (for some fixed theory) box.
Can you find a modal propositional formula expressing the consistency
of that theory. You know already that <>p will mean that p is
consistent with that theory (adding p will not lead to a
contradiction), so you can express that p is consistent with that
theory, it is <>p.
But how to express the simple consistency of the theory? What does it
mean, in classical logic, that a theory is consistent?
Bruno
http://iridia.ulb.ac.be/~marchal/
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