$vx = z$
$zb = y$
Which means $vxb = y$
Does this build on an axiom (and which)? I have to prove a statement using only some specific axioms. But I don't know if I'm allowed to do that.
$vx = z$
$zb = y$
Which means $vxb = y$
Does this build on an axiom (and which)? I have to prove a statement using only some specific axioms. But I don't know if I'm allowed to do that.
If:
then, yes. If $vx=z$ and $zb=y$, then $y = zb = (vx)b = vxb$. You are using the fact that the product is a function, so evaluating at $z$ and $b$ is the same as evaluating at $vx$ and $b$ (since $vx=z$; this is sometimes called the "Principle of Substitution", which is an axiom of the underlying logic). So $zb= (vx)b$. And because the operation is assumed to be associative, then the two possible meanings of "$vxb$" (namely, $(vx)b$ and $v(xb)$) have the same value, so we do not need to distinguish between them.
In particular, if these are, for instance, real numbers and juxtaposition is the usual multiplication of real numbers, then the implication is valid.