I was doing an AMC question that used a binary operation, ?, that was defined as such:
a?(b?c) = (a?b)*c , for all real, non-zero a,b,c, where * is normal multiplication. In addition, (a?a)=1 for all real, non-zero a.
Using the substitution b=c, we can show:
a?(b?b) = (a?b)*b
a?(a?a) = (a?b)*b
(a?a)*a = (a?b)*b
a = (a?b)*b
leading to the explicit form a?b = a/b.
My question is, why are we allowed to just substitute b=c? Shouldn’t the explicit form only be true when b=c? This may be a stupid question but it got me wondering.
The idea is that the $?$ symbol in $a?b$ represents a binary operation. A binary operation looks at exactly two input values in order to determine its output value, without regard for any other information.
If you needed to know whether $b=c$ in order to say what the value of $a?b$ is, then $a?b$ wouldn't be a binary operation.
But I would approach the problem a bit differently. There's nothing magical about the choice of variables $a$, $b$, or anything else, so let's write $x?y$ and try to figure out how to find the value of $x?y$ using known operations.
Step one: suppose $x\neq 0$ and $y\neq 0.$ Whatever $x?y$ is, multiply it by $y$. Then we have $$ (x?y) \times y. \tag1 $$
According to the definition of $?$, which says that $a?(b?c) = (a?b)\times c$ for all real, non-zero $a,b,c$, we can match the variables $a,b,c$ in that definition with variables in formula $(1)$, and so we know that $$ (x?y) \times y = x?(y?y). \tag2 $$
But the definition also tells us that $y?y = 1,$ so we can simplify the right-hand side of Equation $(2)$: $$ (x?y) \times y = x?1. \tag3 $$
Now let's write another formula (not derived from any of the ones in $(1)$, $(2)$, or $(3)$) using just $x$: $$ x?(x?x). \tag4 $$ This can be simplified in two different ways using the fact that $x?x=1$: $$ x?(x?x) = x?1. \tag5 $$ $$ x?(x?x) = (x?x) \times x = 1 \times x = x. \tag6 $$
Since the left-hand sides of Equations $(5)$ and $(6)$ are equal (the value is $x?(x?x)$ each time), the right-hand sides also are equal:
$$ x?1 = x. \tag7 $$
Combine Equations $(3)$ and $(7)$: $$ (x?y) \times y = x. \tag8 $$
Now since we originally supposed that $y\neq 0,$ we can divide by $y$ on both sides of Equation $(8)$: $$ x?y = \frac xy. $$
This is a somewhat more long-winded version of the proof you cited in the question. One stylistic difference is that in my proof, I never say I am about to substitute one variable for another (such as by letting $b = c$); I simply write the same variable name twice in the same expression, $(x?y) \times y.$ Since the definition of $?$ says nothing about the three values $a$, $b$, and $c$ being distinct from each other, I am allowed to write such things and to use the definition of $?$ to evaluate them.
A major benefit of the "just write it" technique is that I can write $x?(x?x)$ without having to explain what variable is being substituted for what. Imagine trying to explain how we get to $a?(a?a)$ (as required by your proof) by substitution; do we say that $a = b$ and $a = c$? That indeed would cause a problem when we then claimed we had found a general expression for $a?b$.
So your proof is not wrong, but on the other hand you are not wrong to question why it says $b = c$. It is, in fact, completely unnecessary to "make the substitution $b = c$", and I would argue that taking an unnecessary step like that is actually harmful.