Re: Semigroup problem

On Sun, 11 May 2008 marcjhg@xxxxxxxxx wrote:

Assume we work in first-order logic with equality. I will use SallT
for the universal quantifier and SexistsT for the existential
quantifier. The symbol != denotes inequality. F and G are functions of
arity 1. * is function of arity 2. I will use infix notation for *. (%
will be a symbol for comment)

In the lines above and below, what's the meanings of
SallT
SexistsT
Sweak
communitivityT
itRs

The axioms are:

all x all y exists z (x * y = z). % Closure
all x all y all z (x * (y * z) = (x * y) * z). % Associativity
all z exists x exists y (x * y = z). % Surjectivity
all x all y (F(x) * G(y) = y * x). % SWeak
commutativityT

Assume semigroup with identity e and left and right cancellation.

f(e) g(y) = y
f(x) g(e) = x

f(e) g(e) = e = ee = f(e) g(e) f(e) g(e)
e = g(e) f(e)

xy = f(x) g(e) f(e) g(y) = f(x) g(y) = yx

On the ground of these axioms only try to prove:
all x all y (x * y = y * x). % Commutativity

Or try to find a model (finite if itRs possible) in which holds:
exists x exists y (x * y != y * x). % Non-commutativity

.

Relevant Pages

  • Re: Semigroup problem
    ... for the universal quantifier and SexistsT for the existential ... The symbol!= denotes inequality. ... and he can only see ASCII characters, ... Yes, this implies commutativity. ...
    (sci.math)
  • Re: Semigroup problem
    ... for the universal quantifier and SexistsT for the existential ... The symbol!= denotes inequality. ... Assume semigroup with identity e and left and right cancellation. ... Yes, this implies commutativity. ...
    (sci.math)