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Given this function specification, where name xs is bound to a list, # denotes its cardinality and . is the list index operator...

$$\left( \sum i: 0 \leq i < \#xs: xs.i * (i + 1) \right)$$

I need to derive a recursive function using induction.

Base case: []

\begin{align} \left( \sum i: 0 \leq i < \#[]: xs.i * (i + 1) \right) && \text{(Textual substitution - xs is free)} \\ \left( \sum i: 0 \leq i < 0: xs.i * (i + 1) \right) && \text{(Def. of #)} \\ \left( \sum i: False: xs.i * (i + 1) \right) && \text{(Algebra)} \\ \left( \sum i: 0 \leq i < \#xs: xs.i * (i + 1) \right) && \text{(Empty range)} \end{align}

Inductive case: (x:xs)

\begin{align} \left( \sum i: 0 \leq i < \#(x:xs): (x:xs).i * (i + 1) \right) && \text{(Textual substitution - xs is free)} \\ \left( \sum i: 0 \leq i < 1 + \#xs: (x:xs).i * (i + 1) \right) && \text{(Def. of #)} \end{align}

How can I proceed from here ?

F. Zer
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1 Answers1

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Your notation and understanding are pretty good.

It is easier to consider (xs:x) as the inductive case instead of (x:xs)

\begin{align} \sum_{i:\ 0 \leq i < \#(xs:x)} &(xs:x).i * (i + 1)\\ &=\sum _{i:\ 0 \leq i < \#xs+1} (x:xs).i * (i + 1) \\ &=\sum_{i:\ 0 \leq i < \#xs} (xs:x).i * (i + 1)+\sum_{i:\ i=\#xs}(xs:x).i * (i + 1)\\ &=\sum_{i:\ 0 \leq i < \#xs} xs.i * (i + 1)+x*(\#xs+1),\\ \end{align} where we assume that list index starts with 0.

If we denote the function by $f$, the above equality becomes $$ f(xs:x) = f(xs) + x*(\#xs+1), $$ which is the recursive step of a recursive definition.

The base case, as you have indicated, is $$ f([]) = \sum_{ i:\ 0 \leq i < 0} [].i * (i + 1) = \sum_{i: \text{ empty set}}[].i * (i + 1) =0.$$

Exercise. Derive the following recursive formula of the same function $f$. $$ f(x:xs) = f(xs) + t(xs)+ x, $$ where $$t(xs)=\sum_{i:\ 0 \leq i < \#(xs)} xs.i,$$ the sum of items in $xs$.

John L.
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