Originally Posted by
xalk No ,no i mean what axioms for the order.For example one such system is the following:
1) $\displaystyle \forall x\forall y[ x<y\Longrightarrow \neg (y<x)]$
2) $\displaystyle \forall x\forall y\forall z[x<y\wedge y<z\Longrightarrow x<z]$
3)$\displaystyle \forall x\forall y[ x\neq y\Longrightarrow (x<y\vee y<x)]$
4)$\displaystyle \forall x\forall y\forall z[y<z\Longrightarrow (x+y<x+z)]$
5)$\displaystyle \forall x\forall y\forall z[0<x\wedge y<z\Longrightarrow x.y<x.z]$
Now in a such axiomatic system, if in axiom (5) you put:
x= a , y= 0 and z=b we have the following result:
$\displaystyle 0< a\wedge 0< b\Longrightarrow 0.a< a.b$ ,which is exactly the theorem you are trying to prove ,since 0.a = 0
However in another axiomatic system for order we can have a different proof