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$vx = z$

$zb = y$

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.

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It would be very useful to the question to say which axioms you are allowed to use, and what is the context of the multiplication. –  Asaf Karagila Sep 20 '11 at 16:25
    
Also, is this homework? If this is homework please add [homework] tag to the question. –  Asaf Karagila Sep 20 '11 at 16:25

1 Answer 1

up vote 2 down vote accepted

If:

  1. Juxtaposition represents the result of performing a binary function written in infix notation; and
  2. This binary function is associative (so that $(ab)c = a(bc)$ for all $a$, $b$, and $c$;

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.

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