Datatype Rules

Datatype rules may seem a bit difficult to understand at first sight. Here are a few examples intended to make them clearer.

Datatypes used

The rules will be applied to the following datatypes:

Directions:

direction ::=
    north
  | east
  | south
  | west

Let us have a function

$\operatorname{move} \in \Z \cprod \Z \cprod \operatorname{direction} \tfun \Z \cprod \Z$

first two arguments are a start position, third is a direction, result is the end position.

Lists:

list(T) ::=
    nil
  | cons( head : T, tail : list(T) )

Let us have a predicate

$\operatorname{member}(T, list(T))$

meaning that the first argument is a member of the list.

Trees:

tree(T) ::=
  empty
| node( left : tree(T), value : T, right : tree(T) )

Let us have a function

$\operatorname{height} \in \operatorname{Tree} \tfun \N$ defining the height of a tree

Distinct Case

The rule states

$\frac{\textbf{H}, x=c_1(p_{11}, \ldots, p_{1k}) \;\;\vdash \;\; \textbf{G} \qquad \ldots \qquad \textbf{H}, x=c_n(p_{n1}, \ldots, p_{nl}) \;\;\vdash \;\; \textbf{G} }{ \textbf{H} \;\;\vdash\;\; \textbf{G} }$

Application to directions

Starting from goal

$\forall x,y,d \qdot \operatorname{move}(x \mapsto y \mapsto d) \neq x \mapsto y$

we can free variables x,y,d then apply Distinct Case:

The resulting proof step is

$\frac{ d=\operatorname{north} \;\vdash \; \operatorname{move}(x \mapsto y \mapsto \operatorname{north}) \neq x \mapsto y  \qquad  d=\operatorname{east} \;\vdash \; \operatorname{move}(x \mapsto y \mapsto \operatorname{east}) \neq x \mapsto y \qquad  d=\operatorname{south} \;\vdash \; \operatorname{move}(x \mapsto y \mapsto \operatorname{south}) \neq x \mapsto y \qquad  d=\operatorname{west} \;\vdash \; \operatorname{move}(x \mapsto y \mapsto \operatorname{west}) \neq x \mapsto y }{  \;\;\vdash\;\; \operatorname{move}(x \mapsto y \mapsto d) \neq x \mapsto y }$

further simplifications depend on the existence of rules about move and the various directions.

Application to lists

Starting from goal

$\forall l \qdot \exists x \qdot \operatorname{member}(x,l) \land (\forall y \qdot \operatorname{member}(y,l) \limp y \leq x)$

we can free variable l then apply Distinct Case on it:

$\frac{ l=\operatorname{nil} \;\vdash \; \exists x \qdot \operatorname{member}(x,\operatorname{nil}) \land (\forall y \qdot \operatorname{member}(y,\operatorname{nil}) \limp y \leq x)  \qquad  l=\operatorname{cons}(head0, tail1) \;\vdash \; \exists x \qdot \operatorname{member}(x,\operatorname{cons}(head0, tail1)) \land (\forall y \qdot \operatorname{member}(y,\operatorname{cons}(head0, tail1)) \limp y \leq x) }{  \;\;\vdash\;\; \exists x \qdot \operatorname{member}(x,l) \land (\forall y \qdot \operatorname{member}(y,l) \limp y \leq x) }$

The left antecedent shows that the case of the nil list has been forgotten when writing the theorem.

Application to trees

TODO

!l,v,r. t = node(l,v,r) => height(l) < height(t)

Induction

TODO