Re: a^b^c ?
- From: "rio" <a@xxx>
- Date: Thu, 20 Nov 2008 18:16:14 +0100
"rio" <a@xxx> ha scritto nel messaggio
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"Herbert Kleebauer" <klee@xxxxxxxxx> ha scritto nel messaggio
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rio wrote:
there is someone that know the logic [or the mathematic] can answer this?
Another operator which usually associates to the right is logical
implication: a => b => c is mostly meant to mean a => (b => c).
Actually, I would expect a => b => c to mean (a => b) => c
for the same reason you gave above, i.e. because for a => (b => c)
there is the simpler notation (a & b) => c.
Yet another meaning of a => b => c could be (a => b) and (b => c),
especially when used in an informal way where no formal logic is expected.
Implications can in general have many meanings. a => b can mean
the problem born that
in mathematic theorem demonstration someone use to write
a => b => c => d => ...
what "a => b => c" means: it means "a => (b => c)"
or it means "(a => b) => c"
i show if the proposition "a" is true ["a" is in the hypothesis of the theorem
so must be true]
than write "a => (b => c)" is equivalent to write "(a => b) => c"
so it is not important put parenthesis
for say the true i wrote a big copybook [quaderno]
of all these formulas all are in relation each other
so one have to choice some ones like axioms
it is all like algebra but all formal write down
follow allow formulas
this is what i think... :)
eg. a==(13eN) true
[(a => b) => c] <=> [a => (b => c)]
Dim
a is true
(a/\¬b) <=> ¬b <=> (¬aV¬b)
so
[(a => b) => c] <=> ¬(¬aVb)Vc <=> (a/\¬b)Vc
<=> (¬aV¬b)Vc <=> ¬aV(¬bVc) <=> [a => (b => c)]
the theorem to show is this not the above the theorem is
a=> {[(a => b) => c] <=> [a => (b => c)]}
all math is formal
.
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