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I have been reading Economics and Consumer Behavior by Angus Deaton and John Muellbauer, specifically reading up on Composite Commodity Theorem, which states:

if prices move in parallel to each other, then the corresponding group of commodities can be treated as a single good. (page 120 in the book)

This is a pretty mind blowing result (well at least for me), however im yet to find an actual proof for the theorem in the literature.

is there a proof for this result?

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    $\begingroup$ @Giskard Whoopsie. $\endgroup$ – EconJohn Jun 28 at 16:56
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There is this nice expository note:
Carter M., 1995, "An expository note on the composite commodity theorem," Economic Theory, 5, 175-179
and an interesting generalization by
Lewbel, A., 1996, "Aggregation without Separability: A Generalized Composite Commodity Theorem," American Economic Review, 86, 524-543.

EDIT:
One reason why it is difficult to find a proof of the theorem, is that the so called "composite commodity theorem" does not really have the mathematical status of a theorem. It is more a principle directly following from the reparameterization below.
Let $x\in \mathbb{R}^J_+$ denotes the disaggregate demand system, which depends on the price vector $p \in \mathbb{R}^J_+$, and the budget $b>0$. The relative prices are denoted by $\alpha = p/\textbf{p}$ where the aggregate price index is $\textbf{p} \in \mathbb{R_+}$. If $\alpha $ is a vector of constants, we can simplify the demand system such that it depends only upon the aggregate price $\bf{p}$: $$ x(p,b) = x(\alpha \mathbf{p},b) \equiv \mathbf{x}(\mathbf{p},b). $$ Note that now $\mathbf{x}:\mathbb{R^2_+} \rightarrow \mathbb{R}^J_+$. If one is also interested in aggregating the elementary quantities, it is (for instance) possible to define the aggregate composite quantity as: $$ \mathbf{X}(\mathbf{p},b) = \alpha^T\mathbf{x}(\mathbf{p},b), $$ where $ \mathbf{X}:\mathbb{R^2_+} \rightarrow \mathbb{R}_+$. In this case, the elementary and aggregate expenditures are the same, in the sense that: $$\mathbf{p}\mathbf{X}(\mathbf{p},b)=p^Tx(p,b).$$

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