Structural induction is used to prove that some proposition P(x) holds for all x of some sort of recursively defined structure, such as formulas, lists, or trees. A well-founded partial order is defined on the structures ("subformula" for formulas, "sublist" for lists, and "subtree" for trees).

We prove that P(x) is true for all x ∈ S by structural induction. Inductive Step: Goal: Show P(x+y) Since P(x) is true, 3|x and so x=3m for some integer m and since P(y) is true, 3|y and so y=3n for some integer n. Therefore x+y=3m+3n=3(m+n) and …

Structural induction A method for proving properties of recursive structures. Using structural induction Example proofs about recursively defined numbers, strings, and trees. 2

We prove that P(x) is true for all x ∈ S by structural induction. Inductive Step: Goal: Show P(x+y) Since P(x) is true, 3|x and so x=3m for some integer m and since P(y) is true, 3|y and so y=3n for some integer n. Therefore x+y=3m+3n=3(m+n) and thus 3|(x+y). Hence P(x+y) is true. Claim: Every element of is divisible by 3. Let P(x) be “3|x”.

2020年5月18日 · Structural induction is used to prove that some proposition P (x) holds for all x of some sort of recursively defined structure, such as formulae, lists, or trees—or recursively- defined sets.

Proofs by Structural Induction • Extends inductive proofs to discrete data structures -- lists, trees, … • For every recursive definition there is a corresponding structural induction rule.

We prove by structural induction that P(T) holds for every binary tree. Base case: If T = , by de nition jnodes(T)j= 0 and h(T) = 1. jnodes(T)j= 0 = 20 1 = 2 1+1 1 = 2h(T)+1 1 (1) If T 2V (T is a tree consisting of a single node), it is easy to see that h(T) = 0 and jnodes(T)j= 1. Then jnodes(T)j= 1 = 21 1 = 20+1 1 = 2h(T)+1 1 (2)

Lecture 21: Structural induction. Reading: MCS 7,7.1. Proofs by structural induction. Review exercises: Prove that \(len(cat(x,y)) = len(x) + len(y)\). Prove that \(len(reverse(x)) = len(x)\). Use the inductive definitions of \(\mathbb{N}\) and \(plus\) to show that \(plus(a,b) = plus(b,a)\). Idea behind structural induction

Structural Induction is the most general form of induction. In Structural Induction, we can have multiple object constants, multiple function constants, and, unlike our other forms of induction, we can have function constants with multiple arguments.

We can use induction to prove properties of recursively defined objects. This is called structural induction. Theorem: Properly nested strings of left and right parentheses are balanced. A string of parentheses x is called balanced if left[x] = right[x], where left[x] (right[x]) is the number of left (right) parentheses in x.