CSE 311 Lecture 19: Structural Induction

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Topics

Recursively defined sets: A brief review of Lecture 18.
Structural induction: A method for proving properties of recursive structures.
Using structural induction: Example proofs about recursively defined numbers, strings, and trees.

Recursively defined sets

A brief review of Lecture 18.

Giving a recursive definition of a set

A recursive definition of a set $S$ has the following parts:: Basis step specifies one or more initial members of $S$.; Recursive step specifies the rule(s) for constructing new elements of $S$ from the existing elements.; Exclusion (or closure) rule states that every element in $S$ follows from the basis step and a finite number of recursive steps.

The exclusion rule is assumed, so no need to state it explicitly.

Recursively strings and functions on them

Let $\Sigma$ be a finite set of characters, and define $\Sigma^*$ to be the set of of all strings over $\Sigma$:: Basis: $\varepsilon \in \Sigma^* $, where $\varepsilon$ is the empty string.; Recursive: if $w\in\Sigma^* $ and $a\in\Sigma$, then $wa\in\Sigma^* $
Length: $\op{len}(\varepsilon) = 0$; $\op{len}(wa) = \op{len}(w) + 1$ for $w\in\Sigma^* $, $a\in\Sigma$
Reversal: $\varepsilon^{\op{R}} = \varepsilon$; $(wa)^{\op{R}} = aw^{\op{R}}$ for $w\in\Sigma^* $, $a\in\Sigma$
Concatenation: $x\bullet \varepsilon = x$ for $x\in\Sigma^* $; $x\bullet (wa) = (x\bullet w)a$ for $x,w\in\Sigma^* $, $a\in\Sigma$
Number of $c$’s in a string: #$_ c(\varepsilon) = 0$; #$_ c(wc) = $ #$_ c(w) + 1$ for $w\in\Sigma^* $; #$_ c(wa) = $ #$_ c(w)$ for $w\in\Sigma^* $, $a\in\Sigma$, $a\neq c$

Rooted binary trees and functions on them

Rooted binary trees: Basis: $\Node\in S$; Recursive: if $L\in S$ and $R\in S$, then $\Tree(\Node, L, R) \in S$
Size of a rooted binary tree: $\Size{\Node} = 1$; $\Size{\Tree(\Node, L, R)} = 1 + \Size{L} + \Size{R}$
Height of a rooted binary tree: $\Height{\Node} = 0$; $\Height{\Tree(\Node, L, R)} = 1 + \max(\Height{L}, \Height{R})$

Structural induction

A method for proving properties of recursive structures.

How can we prove properties of recursive structures?

Suppose that $S$ is a recursively defined set.: And we want to prove that every element of $S$ satisfies a predicate $P$.
Can we use ordinary induction to prove $\forall x\in S. P(x)$?: Yes! Define $Q(n)$ to be “for all $x\in S$ that can be constructed in at most $n$ recursive steps, $P(x)$ is true.”

But this proof would be long and cumbersome to do!
So we use structural induction instead.
- Follows from ordinary induction (on $Q$), while providing a more convenient proof template for reasoning about recursive structures.
- As powerful as ordinary induction, which is just structural induction applied to the recursively defined set of natural numbers.

Proving $\forall x\in S. P(x)$ by structural induction

① Let $P(x)$ be [ definition of $P(x)$ ].: We will show that $P(x)$ is true for every $x\in S$ by structural induction.
② Base cases:: [ Proof of $P(s_0), \ldots, P(s_m)$. ]
③ Inductive hypothesis:: Assume that $P(y_0), \ldots, P(y_k)$ are true for some arbitrary $y_0, \ldots, y_k \in S$.
④ Inductive step:: We want to prove that $P(y)$ is true.; [ Proof of $P(y)$. The proof must invoke the structural inductive hypothesis. ]
⑤ The result follows for all $x\in S$ by structural induction.

Recursive definition of $S$: Basis step: $s_0\in S, \ldots, s_m\in S$.; Recursive step:; if $y_0, \ldots, y_k\in S$, then $y\in S$.

If the recursive step of $S$ includes multiple rules for constructing new elements from existing elements, then
③ assume $P$ for the existing elements in every rule, and
④ prove $P$ for the new element in every rule.

Structural induction works just like ordinary induction

① Let $P(x)$ be [ definition of $P(x)$ ].: We will show that $P(x)$ is true for every $x\in \N$ by structural induction.
② Base cases:: [ Proof of $P(0)$. ]
③ Inductive hypothesis:: Assume that $P(n)$ is true for some arbitrary $n \in \N$.
④ Inductive step:: We want to prove that $P(n+1)$ is true.; [ Proof of $P(n+1)$. The proof must invoke the structural inductive hypothesis. ]
⑤ The result follows for all $x\in \N$ by structural induction.

Recursive definition of $\N$: Basis step: $0 \in \N$.; Recursive step:; if $n\in \N$, then $n+1\in \N$.

Ordinary induction is just structural induction applied to the recursively defined set of natural numbers!

Understanding structural induction

$\rule{P(\Node); \forall L, R\in S. (P(L)\wedge P(R))\rightarrow P(\Tree(\Node,L,R))}{\forall x \in S. P(x)}$

How do we get $P(\Tree(\Node,\Node,\Tree(\Node, \Node,\Node)))$ from $P(\Node)$ and $\forall L,R\in S. (P(L)\wedge P(R))\rightarrow P(\Tree(\Node,L,R))$?

Define $S$ by: Basis: $\Node \in S$.; Recursive:; if $L, R\in S$, then; $\Tree(\Node,L,R)\in S$

1.	First, we have $\forall L,R\in S. (P(L)\wedge P(R))\rightarrow P(\Tree(\Node,L,R))$
2.	Next, we have $P(\Node)$.	$P(\Node)$
3.	Intro $\wedge$ on 2 gives us $P(\Node)\wedge P(\Node)$.	$P(\Node)\wedge P(\Node)$
4.	Elim $\forall$ on 1 gives us $(P(\Node)\wedge P(\Node))\rightarrow P(\Tree(\Node, \Node, \Node))$.	$\ \Downarrow_{\ (P(\Node)\wedge P(\Node))\rightarrow P(\Tree(\Node, \Node,\Node))}$
5.	Modus Ponens on 3 and 4 gives us $P(\Tree(\Node, \Node,\Node))$.	$P(\Tree(\Node, \Node,\Node))$
6.	Intro $\wedge$ on 2 and 5 gives us $P(\Node)\wedge P(\Tree(\Node, \Node,\Node))$.	$P(\Node)\wedge P(\Tree(\Node, \Node,\Node))$
7.	Elim $\forall$ on 1 gives us $(P(\Node)\wedge P(\Tree(\Node, \Node,\Node))\rightarrow P(\Tree(\Node,\Node,\Tree(\Node, \Node, \Node)))$.	$\ \Downarrow_{\ (P(\Node)\wedge P(\Tree(\Node, \Node,\Node))\rightarrow P(\Tree(\Node,\Node,\Tree(\Node, \Node, \Node)))}$
8.	Modus Ponens on 6 and 7 gives us $P(\Tree(\Node,\Node,\Tree(\Node, \Node, \Node)))$.	$P(\Tree(\Node,\Node,\Tree(\Node, \Node, \Node)))$

Using structural induction

Example proofs about recursively defined numbers, strings, and trees.

Prove that every $x\in S$ is divisible by 3

① Let $P(x)$ be $3 \vert x$.: We will show that $P(x)$ is true for every $x\in S$ by structural induction.
② Base cases ($x=6$, $x=15$):: $3 \vert 6$ so $P(6)$ holds, and $3 \vert 15$ so $P(15)$ holds.
③ Inductive hypothesis:: Assume that $P(x), P(y)$ are true for some arbitrary $x,y \in S$.
④ Inductive step:: We want to prove that $P(x+y)$ is true.; By the inductive hypothesis, $3\vert x$ and $3\vert y$, so $x = 3i$ and $y = 3j$ for some $i,j\in\Z$. Therefore, $x + y = 3i + 3j = 3(i+j)$ so $3\vert (x+y)$. Hence, $P(x+y)$ is true.
⑤ The result follows for all $x\in S$ by structural induction.

Define $S$ by: Basis: $6\in S$, $15\in S$.; Recursive: if $x,y\in S$, then $x+y\in S$.

Prove $\op{len}(x\bullet y) = \op{len}(x) + \op{len}(y)$ for all $x,y\in\Sigma^* $

What object ($x$ or $y$) to do structural induction on?

① Let $P(y)$ be $\forall x\in\Sigma^* . \op{len}(x\bullet y) = \op{len}(x) + \op{len}(y)$.: We will show that $P(y)$ is true for every $y\in \Sigma^* $ by structural induction.
② Base case ($y=\varepsilon$):: For every $x\in \Sigma^* $, $\op{len}(x\bullet \varepsilon)$ $=$ $\op{len}(x)$ $=$ $\op{len}(x) + \op{len}(\varepsilon)$ since $\op{len}(\varepsilon) = 0$. So $P(\varepsilon)$ is true.
③ Inductive hypothesis:: Assume that $P(w)$ is true for some arbitrary $w \in \Sigma^* $.
④ Inductive step:: We want to prove that $P(wa)$ is true for every $a\in\Sigma$.; Let $a\in\Sigma$ and $x\in\Sigma^* $ be arbitrary. Then; $\begin{align} \op{len}(x\bullet wa) &= \op{len}((x\bullet w)a) && \text{by defn of } \bullet\\ &= \op{len}(x\bullet w) + 1 && \text{by defn of } \op{len}\\ &= \op{len}(x) + \op{len}(w) + 1 && \text{by IH}\\ &= \op{len}(x) + \op{len}(wa) && \text{by defn of } \op{len}\\ \end{align}$; So $\op{len}(x\bullet wa)=\op{len}(x) + \op{len}(wa)$ for all $x\in\Sigma^* $, and $P(wa)$ is true.
⑤ The result follows for all $y\in \Sigma^* $ by structural induction.

Define $\Sigma^* $ by: Basis: $\varepsilon \in \Sigma^* $.; Recursive:; if $w\in\Sigma^* $ and $a\in\Sigma$,; then $wa\in\Sigma^* $
Length: $\op{len}(\varepsilon) = 0$; $\op{len}(wa) = \op{len}(w) + 1$
Concatenation: $x\bullet \varepsilon = x$; $x\bullet (wa) = (x\bullet w)a$

Prove $\Size{t}\leq 2^{\Height{t}+1}-1$ for every rooted binary tree $t$

① Let $P(t)$ be $\Size{t}\leq 2^{\Height{t}+1}-1$.: We will show that $P(t)$ is true for every $t\in S $ by structural induction.
② Base case ($t=\Node$):: $\Size{\Node} = 1 = 2^1 - 1 = 2^{0+1}-1 = 2^{\Height{\Node}+1}-1$ so $P(\Node)$ is true.
③ Inductive hypothesis:: Assume that $P(L)$ and $P(R)$ are true for some arbitrary $L, R \in S$.
④ Inductive step:: We want to prove that $P(\Tree(\Node,L,R))$ is true.; $\begin{align} \Size{\Tree(\Node,L,R)} &= 1 + \Size{L} + \Size{R} && \text{by defn of } \Size{}\\ &\leq 1 + (2^{\Height{L}+1}-1) + (2^{\Height{R}+1}-1) && \text{by IH}\\ &\leq 2^{\Height{L}+1} + 2^{\Height{R}+1} - 1 && \text{algebra}\\ &\leq 2^{\max(\Height{L},\Height{R})+1} + 2^{\max(\Height{L},\Height{R})+1} - 1 && \text{by defn of } \max \\ &\leq 2(2^{\max(\Height{L},\Height{R})+1}) - 1 && \text{algebra} \\ &= 2(2^{\Height{\Tree(\Node,L,R)}}) - 1 && \text{by defn of }\Height{} \\ &= 2^{\Height{\Tree(\Node,L,R)}+1} - 1 && \text{which is the desired result.} \\ \end{align}$
⑤ The result follows for all $t\in S$ by structural induction.

Define $S$ by: Basis: $\Node \in S$.; Recursive:; if $L, R\in S$, then; $\Tree(\Node,L,R)\in S$
Size: $\Size{\Node} = 1$; $\Size{\Tree(\Node,L,R)} = $
$\quad 1 + \Size{L} + \Size{R}$
Height: $\Height{\Node} = 0$; $\Height{\Tree(\Node,L,R))} = $
$\quad 1 + \max(\Height{L}, \Height{R})$

Summary

To define a set recursively, specify its basis and recursive step.: Recursive set definitions assume the exclusion rule.; We use recursive functions to operate on elements of recursive sets.
Use structural induction to prove properties of recursive structures.: Structural induction follows from ordinary induction but is easier to use.
To prove $\forall x\in S. P(x)$ using structural induction:: Show that $P$ holds for the elements in the basis step of $S$.; Assume $P$ for every existing element of $S$ named in the recursive step.; Prove $P$ for every new element of $S$ created in the recursive step.