Hi Liz,
May be I am to quick.
On 22 Jan 2014, at 12:58, Bruno Marchal wrote:
On 22 Jan 2014, at 04:23, LizR wrote:
On 21 January 2014 22:29, Bruno Marchal <[email protected]> wrote:
No, it is all good, Liz!
What about:
(p V q) -> p
Using the same formula this is equivalent to(~(p V q) V p), which
for (0,1) is 0, hence not a law.
and
p -> (p & q)
And this is (~p V (p & q)) which is 0 for (1,0), hence also not a
law :-)
What about (still in CPL) the question:
is (p & q) -> r equivalent with p -> (q -> r)
is (~(p & q) V r) equal to (~p V (~q V r)) ?
or is
~((p & q) & ~r) equal to ~(p & ~~(q & ~r))
i.e. is
~((p & q) & ~r) equal to ~(p & (q & ~r))
I'm going to take a punt and assume the order in which things are
ANDed together doesn't matter, in which case the above comes out as
equal (equivalent). Did I blow it?
Not sure that I understand what you mean by blowing it. But you are
correct in all answers.
Oh! You did not answer:
((COLD & WET) -> ICE) -> ((COLD -> ICE) V (WET -> ICE))
Eek! That is even more difficult. Luckily you provided something
that didn't involve so much typing...
((p & q) -> r) -> ((p -> r) V (q -> r))
Expanding furiously and trying not to make any mistakes...
~((p & q) & ~r) -> ( ~(p & ~r) V ~(q & ~r))
~(~((p & q) & ~r) & ~(~(p & ~r) V ~(q & ~r)))
Um! Assuming for a moment that's correct, we have 8 possible
combinations of values for p,q,r
r = 0 gives 1 (so that's half the values sorted)
r = 1 also gives 1 (so that's the other half)
So assuming I expanded it correctly, it's a law.
Very good. And so
((COLD & WET) -> ICE) -> ((COLD -> ICE) V (WET -> ICE))
Is true in all worlds!
Of course it is non sense if you interpret the arrow in "p -> q" as
a causal implication. I use that formula to explain that "->" is not
a causal relation.
I will try more later...
OK.
Oh, it looks we are later:
Actually, you will have to remind me what [] and <> mean before I
go any further.
OK.
Let us take only 3 propositional variable, or letters, in our
language; p, q, and r, say.
A world (in that context) is given when we say which (atomic)
proposition is true, and which is false.
So, with three propositional variable we get 8 worlds, in the
multiverse associated to {p, q, r} (our language).
They are the one in which p, q, and r are all true, the one in which
p, and q are true, but r is false, ..., the one in which p, q and r
are all false.
OK?
If we fix the order p, q, r on {p, q, r}, we can represent a world
by a sequence of 0 and 1 (which by the way are often used to
represent true and false)? The 8 worlds of the multiverse are given by
000
100
010
110
101
001
111
011
OK?
Let A, B, C range over arbitrary formula. I recall that there are
two kind of formula. the atomic formula and the compound formula.
(A, B, C are metavariable. A->B is NOT a formula, unless A and B
designate some formula (which can contains only the formal p, q, r
(and the logical symbols, parentheses, etc.)
It is the same in algebra. x is not number, unless x designate some
number, like when x = 42. OK?
An atomic formula is just a letter from our set of propositional
letter. We call it an atomic proposition when we think about it in
the company of some truth or false assignment (a proposition can be
said true, or false, not a letter!). OK?
CPL is truth-functional. It means that the truth value (in some
world, thus) of a compound formula is determined by the the truth
value of its subformula, that is eventually by its atomic components.
So the semantic here is very easy, and can be described by the truth
tables, where the truth value of a compound formula is put under the
main connector of the formula :
~ p
0 1
1 0
p & q
1 1 1
0 0 1
1 0 0
0 0 0
p V q
1 1 1
0 1 1
1 1 0
0 0 0
p -> q (~p V q)
1 1 1
0 1 1
1 0 0
0 1 0
OK?
Now, your question, what does mean []A, for A some formula. And what
does mean <>A
Well, for Leibniz and Aristotle, it means that A has the value true
in all worlds, or A is true in all worlds, or that all worlds (in
the multiverse) satisfy A.
In particular, given that you have shown that (p -> p) is a law,
true in all worlds, we have that
[] (p -> p)
OK? For each CPL laws A, sometime called "tautology", we will have
that []A.
examples are easly derived from you work. We will have
[] (p -> p)
[] (p -> (q -> p))
[] ((p & q) -> r) -> ((p -> r) V (q -> r))
etc.
OK? That is easy. tautologies, that is laws, are universal, so
they are verified or satisfied in all worlds, and so the fact that
they are laws is true in all worlds.
This will remain valid for the more general Kripke semantics so you
might try to remind this:
If A is true in all worlds, then []A is true in all worlds.
OK?
from this, and the semantic of the not ('~'), can you find the
semantics for the diamond <>?
<>p is true if ~[]~p is true, if []~p is false, which means that
there is a world in which p is true.
Unfortunately all this might not seem helpful for a formula which
mix modal compounds with non modal compounds, like
[]p -> p
Here, there is no more truth table available, and so you have to
think. The Leibniz semantic (the only semantic we have defined)
provides all the information to solve the puzzle.
You might reread my explanation for []p -> p.
Let us verify this for our little multiverse corresponding to our
little language with three propositional variables.
The question is "does all worlds satisfy []p -> p ?
We have 8 worlds. So let us look at the first one: the one in which p,
q and r is true.
Well, p is true, there, isn't it? What about []p ? Well, we don't
know. Is that a problem for deciding if []p -> p is true in that
world? Not really. Because an implication (A -> B) can only be false
in the only case where B is false, and A is true. So if p is true, []p
-> p is (trivially) true. So []p -> p is true in the first of the 8
worlds. In fact this has shown that this is true for all the worlds
where p is true at once, as the truth value of "[]p -> p" does not
depend on the truth value of q and r.
Now what about the world where p is false. Can we have []p -> p being
false there?
Again, an implication (A -> B) is false only if A is true and B is
false. Look at the truth table, or remember that (A -> B) is really ~A
V B, or ~(A & ~B). OK?
So "[]p -> p", to be false in those world where p is false, need both
[]p true, and p false. There. OK?
But, wait, if []p is true, in that case we know that p has to be true
in all worlds (that the Leibnizian meaning).
But if p is true in all worlds then p has to be true in this world,
yet, we do have p false! Contradiction.
So if p is false, []p cannot be true! So if p is false in a world, []p
has to be false there, and so []p -> p is really
false -> false, and so is true, in those world where p is false.
As we have seen that []p -> is true in the world where p is true, and
[]p -> p is false in the world where p is false, now we know that []p -
> p is true in all worlds, (because in all worlds p is true, or false).
To sum up.
In the worlds where p is true, []p -> p is trivially true.
In the worlds where p is false, we know directly by Leibniz semantics
that []p has to be false, so []p -> p is also "trivially" true. The
false implies everything you want, as you can deduce again from the
truth table of
"->".
Did this help?
The Leibnizian semantics is very rough. []A really means that A is
absolutely true in all worlds, independently of the worlds and of any
possible interaction and interference between worlds.
This will lead to a last comment below.
Which happens indeed to be a law in this Leibnizian setting.
And the question is now: which among the following are also
Leibnizian laws:
p -> []p
[]p -> [][]p
[]p -> <>p
p -> []<>p
<>p -> []<>p
<>p -> ~[]<>p
[]p & ([](p -> q) ->. []q (sometimes p ->. q is more readable
than (p -> q). The comma makes precise which is the main connector.
[](p -> q) ->. ([]p -> []q)
You can verify or guess the result by looking at each world in the
little 8 worlds multiverse. Keep in mind that "p -> q" is false (in
some world) only when p is true in that world, and q false in that
world. Look at the truth table of "p -> q".
All those formula will play some role in the drama which will
follow! So you might know their name. In the same order they are
called (Triv, 4, D, B, 5, g, FMP, k), which are for
Trivial = (p -> []p), which will appear to be highly non trivial in
comp.
4 = ([]p -> [][]p), which is an horrible name (a number!), and
denotes the main formula in Lewis system (S4)
D = ([]p -> <>p), with D for deontic. It is a common axiom in
deontic logic: if p is obligatory, then p is permitted.
B = p -> []<>p. B for Brouwer, due to an attempt to formalize
intuitionist logic in a classical logic (that attempt failed, but
other will succeed)
5 = <>p -> []<>p. Again a bad naming! It means the main axiom is
lewis fifth system S5.
g = <>p -> ~[]<>p (G for Gödel, do you see why?)
FMP = []p & ([](p -> q) ->. []q (FMP = Formal Modus Ponens)
k = [](p -> q) ->. ([]p -> []q) (k is for Kripke).
Do you see that k and FMP are equivalent (hint: you already verified
this!).
If something is unclear, ask me to explain again, perhaps
differently, or tell me what you miss.
Help yourself with drawings. You can draw the world by "potatoes",
and put the true sentences in each of them.
My own head is boiling hot, but that is not due to modal logic, but
it is due to both (COLD & WET) being true in my world, and (COLD &
WET) -> SICK, in my poor reality!
That's the modus ponens (MP) rule: if A and (A->B) are true in a
world, then B is true in that world.
I use it to define a rational agent. An agent is rational if he is
not irrational, and an agent is irrational if he believes in both A
and in A -> B, yet he denies B.
To sum up:
[]p is true (in a world of a multiverse) if p is true in all worlds
of that multiverse.
<>p is true (in a world of a multiverse) if p is true in at least
one world in that multiverse.
With that semantics, I could also have written equivalently:
[]p is true (in ALL world of a multiverse) if p is true in all worlds
of that multiverse.
<>p is true (in ALL world of a multiverse) if p is true in at least
one world in that multiverse.
That does not make the Leibnizian semantic trivial, or collapsing all
modalities. Can you build a little multiverse in which p -> []p is
refuted. That is, containing at least one world in which p -> []p is
false?
Don't blow your mind too much, and take the time. tell me if this
helped.
Keep in mind that logicians are literalist. Any difficulties here are
only the product of the imagination, and the lack of taking each
explanation literally.
Let give you an exercice/dialog in literal classical logic. The dialog
provides the solutions, as I guess you might be a bit tired.
Is it true that all white unicorns living in my apartment are taller
than 42 km high?
Answer: yes, that is true.
Is it true that all white unicorns living in my apartment are smaller
than 42 km high?
Answer: yes, that is true.
But is that not contradictory?
Yes. The conclusion is. But it is very useful. Indeed, I can prove now
that there is no white unicorn living in my apartment.
The proof uses a reductio ad absurdo. Suppose that there is a white
unicorn living in my apartment. Well, then by above, they would be
both taller and smaller than 42 km, and that is a contradiction.
Of course, I was assuming that we agree on the theory that unicorns
don't exist. In that case:
x is a unicorn -> x is taller than 42 km
is always true, because whatever x means, "x is a unicorn" is always
false, in the worlds where there are no unicorn.
OK. A last little exercise in the same vein, for the night. (coming
from a book by Jeffrey):
Alicia was singing this:
"Everybody loves my baby. My baby loves nobody but me".
Can we deduce from this that everybody loves Alicia?
Bruno
Bruno
http://iridia.ulb.ac.be/~marchal/
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