Let $X$ be a non-empty set, and suppose that $*:X^2\to X$ is a function with the associative property, that is, $$(x*(y*z))=((x*y)*z),$$ for all $x,y,z\in X$, where we write the image under the function of the pair $(a,b)$ in $X^2$ as $(a*b)$. Let $x_1,x_2,...,x_n$ be elements of $X$ and suppose that the brackets and $*$s are inserted into the string of symbols $x_1x_2...x_n$ to give an expression which can be computed using the function $*$ to give an element of $X$. Show that the computations of all such expressions (for the same $x_1x_2...x_n$ in that order) will result in the same element of $X$. [Hints: Use induction on $n$ with the hypothesis that each such expression equals both of $$(x_1*(x_2*(...*(x_{n-1}*x_n)...)))$$ and $$(((...(x_1*x_2)*...)*x_{n-1})*x_n).]$$
I'm very confused about what exactly I'm doing induction on. My attempt is as follows:
First let $n$ be the number of $*$s appearing the the expression. Then, the base case $n=2$ is given to us. That is $((x_1*x_2)*x_3)=(x_1*(x_2*x_3))$, is the same element of $X$. Now suppose that we have the string $x_1x_2...x_n$, and we insert the $*$s and brackets into the string, so that the resulting expression is computable via the function $*$. We suppose that each expression equals both of $$(x_1*(x_2*(...*(x_{n-1}*x_n)...)))$$ and $$(((...(x_1*x_2)*...)*x_{n-1})*x_n).$$ We note that $n$ is at most $n-1$ here. We wish to prove for the case where $n=n$... I'm not sure if what I'm saying is making any sense here.
