Can anyone tell me the name of this proposition, and give a proof it it is straightforward?
$$ \prod_{i\in I}(1+x_i)=\sum_{J\subseteq I}\prod_{j\in J}x_j $$
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By induction:
Let $I'=I\cup\{m\}$.
We have
$$\prod_{i\in I'}(1+x_i)=(1+x_{m})\prod_{i\in I}(1+x_i)=\prod_{i\in I}(1+x_i)+x_{m}\prod_{i\in I}(1+x_i).$$
On the other hand, as the subsets of $I'$ are those of $I$ without and with $x_{m}$,
$$\sum_{J\subseteq I'}\prod_{j\in J}x_j=\sum_{J\subseteq I}\prod_{j\in J}x_j+\sum_{J\subseteq I}\prod_{j\in J\cup\{x_{m}\}}x_j=\sum_{J\subseteq I}\prod_{j\in J}x_j+x_{m}\sum_{J\subseteq I}\prod_{j\in J}x_j.$$
The identity is vacuously true for $I=\emptyset$.
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In the last line how can you write $\sum_{J\subseteq I}\prod_{j\in J\cup{x_{m}}}x_j=x_{m}\sum_{J\subseteq I}\prod_{j\in J}x_j $ ? – tam63 Sep 10 '20 at 09:22
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@tam63: seems obvious from the range of the product, no ? – Sep 10 '20 at 09:25
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Sorry, I also cannot see how the first and second lines of your solution are equal? – tam63 Sep 10 '20 at 09:26
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yes sorry I can see that answer to my first question now – tam63 Sep 10 '20 at 09:26
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@tam63: by the inductive hypothesis. The statement is known true for $I$. – Sep 10 '20 at 09:26