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I have seen the inductive proof and can't make sense of it. I thought it should be easy to show for $4$ variables.

Here is my way of using the associative property to cover all the cases of $4$ variables:

$$a((bc)d)=(a(bc))d=((ab)c)d=(ab)(cd)=a(b(cd))$$

I don't see how all these terms can be shown to be equal in one fell swoop.

Eric Snyder
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user1153980
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    It's hard to tell what sort of answer you're looking for here. In addition, abstract algebra and algebra/precalculus are really, really far away from one another, which makes it harder to tell specifically what you're asking for. Can you clarify? ---- In a proof, you don't have to prove everything at one go. If you prove $A=B$ and $B=C$ (and your system obeys the transitive property) then you have $A=C$. – Eric Snyder Sep 13 '23 at 23:44
  • As explained here, the inductive step pushes ')'s rightward via the rewrite rule $\rm, (xy)z\ \to\ x(yz),,$ terminating with the right-associated normal form where all ')'s are at the right end. In OP this yields

    $$\begin{align}((ab)c)d &= (ab)(cd) = a(b(cd))\[.3em] (a(bc))d &=a((bc)d) = a(b(cd))\end{align}\qquad\qquad$$

    What don't you understand in the linked general proofs?

     – Bill Dubuque Sep 14 '23 at 00:01

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