Re: A problem in first order logic
- From: Jan Burse <janburse@xxxxxxxxxxx>
- Date: Sun, 30 Apr 2006 23:17:47 +0200
Hi
not exists M M[T]=1 & M[~S]=1 iff not exists M M[T u ~S]=1
By cases, by definition and
by the identities:
A iff A & A
A iff B implies not exists M A iff not exists M B.
definition is:
M[T]=1 iff forall A in T M[A]=1
which implies M[A u B]=1 iff M[A]=1 & M[B]=1
for A intersect B=void and M[{A}]=1 iff M[A]=1.
(How is this proofed?)
case 1: ~S not member of T,
then M[T u ~S]=1 iff M[T]=1 & M[~S]=1 by definition.
case 2: ~S element of T,
then M[T u ~S]=1 iff M[T]=1
iff M[T\~S u ~S]=1
iff M[T\~S]=1 & M[~S]=1 by defnition
iff M[T\~S]=1 & M[~S]=1 & M[~S]=1 by identity
iff M[T]=1 & M[~S]=1 by definition
The rest follows from the second identity.
using that cool notation ??
Thanks. Juan.
PD : Where I could find any text about that cool notation ??
Every book about mathematical logic model theory.
Sometimes instead of M[A]=1 one writes M |= A,
and instead of M[A]=0 one writes M |\= A.
But M[.] has more valuation appeal to me than
.. |= .. Sometimes you see also v(A), v(A,M),
[A], [A]_M, etc.. Its all the same idea.
Bye
BTW: M[.] is overloaded, it is defined for
sets of formulas, i.e. T, and single formulas,
i.e. A.
.
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