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. 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.
Bruno
http://iridia.ulb.ac.be/~marchal/
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