Coinduction is dual to induction. It generalises to corecursion.
It is a method of proof which relies on the fact that any two states of the terminal coalgebra for an endofunctor $H$ must be equal if they are indistinguishable under repeated operations of $H$. That is, there are no proper coalgebra quotient objects. Generally, we show the existence of a bisimulation between states of terminal coalgebra, that is a relation between states, such that when the coalgebra function is applied, the respective outputs are still related. Since any bisimulation must be contained within the identity relation, we can then conclude that the states are equal.
Let $\bar{\mathbb{N}}$ be the set of conatural numbers and take $add\colon \bar{\mathbb{N}} \times \bar{\mathbb{N}} \to 1 + \bar{\mathbb{N}} \times \bar{\mathbb{N}}$. We can then establish a bisimulation between the terms $(n + m)$ and $(m + n)$, from which we can conclude that this addition is commutative. (See p. 52 of Rutten Universal coalgebra: a theory of systems.)
Many theorems of calculus involving formal power series $f(x) = \sum_{n \geq 0} \frac{a_n x^n}{n!}$ are naturally viewed as proofs by coinduction. The collection of power series $k[ [x] ]$ over a field $k$ (say of characteristic $0$) may be viewed as a stream coalgebra $k^\mathbb{N}$, i.e., the terminal coalgebra of the endofunctor $k \times -:\; Set \to Set$, with the coalgebra structure
Thus if we set up a bisimulation $\sim$ between two power series $f, g$, establishing $f(0) = g(0)$ and $(D f) \sim (D g)$, we may conclude by coinduction that $f = g$ (else, $k[ [x] ]/\sim$ would be a proper quotient).
As an illustration: consider a field $k$ of characteristic $0$ and, for $r \in k$, define $(1 + x)^r$ to be the power series expansion of $\exp(r \cdot \log(1 + x))$, where $\exp(x) = \sum_{n \geq 0} \frac{x^n}{n!}$ and $\log(1 + x) = \sum_{n \geq 1} (-1)^{n+1}\; \frac{x^n}{n}$ is the inverse of $\exp(x)-1$. To prove the generalized binomial theorem
where $r^\underline{k}$ is the falling power $r(r-1)\ldots (r-k+1)$, we introduce a bisimilarity
and observe first that the constant coefficients of $f(x)$ and $(1+x)^r$ are both $1$, and that
(because $r^{\underline{k+1}} = r \cdot (r-1)^{\underline{k}}$) and similarly
by the chain rule, whence $D(f/r) \sim D((1+x)^r/r)$. Hence $f/r$ is bisimilar to $(1+x)^r/r$, and we conclude in the terminal coalgebra that $f(x) = (1+x)^r$.
Bart Jacobs, Jan Rutten, A tutorial on (Co)Algebras and (Co)Induction (pdf)
Davide Sangiorgi, Introduction to Bisimulation and Coinduction, Cambridge Universtity Press (2012) (web)
Davide Sangiorgi, Jan Rutten (eds.), Advanced Topics in Bisimulation and Coinduction, Cambridge Universtity Press (2012) (web)
Discussion of differential calculus in terms of coinduction is in
About coinduction in cubical type theory:
Last revised on December 16, 2023 at 10:14:12. See the history of this page for a list of all contributions to it.