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138 lines
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138 lines
4.3 KiB
ReStructuredText
This page attempts to explain some of the techniques used in Idris to prove
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propositional equalities.
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Proving Propositional Equality
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==============================
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We have seen that definitional equalities can be proved using ``Refl`` since they
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always normalise to values that can be compared directly.
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However with propositional equalities we are using symbolic variables, which do
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not always normalise.
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So to take the previous example:
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.. code-block:: idris
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plusReducesR : (n : Nat) -> plus n Z = n
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In this case ``plus n Z`` does not normalise to n. Even though both sides of
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the equality are provably equal we cannot claim ``Refl`` as a proof.
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If the pattern match cannot match for all ``n`` then we need to
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match all possible values of ``n``. In this case
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.. code-block:: idris
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plusReducesR : (n : Nat) -> plus n Z = n
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plusReducesR Z = Refl
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plusReducesR (S k)
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= let rec = plusReducesR k in
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rewrite rec in Refl
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we can't use ``Refl`` to prove ``plus n 0 = n`` for all ``n``. Instead, we call
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it for each case separately. So, in the second line for example, the type checker
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substitutes ``Z`` for ``n`` in the type being matched, and reduces the type
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accordingly.
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Replace
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=======
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This implements the 'indiscernability of identicals' principle, if two terms
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are equal then they have the same properties. In other words, if ``x=y``, then we
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can substitute y for x in any expression. In our proofs we can express this as:
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if x=y
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then prop x = prop y
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where prop is a pure function representing the property. In the examples below
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prop is an expression in some variable with a type like this: ``prop: n -> Type``
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So if ``n`` is a natural number variable then ``prop`` could be something
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like ``\n => 2*n + 3``.
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To use this in our proofs there is the following function in the prelude:
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.. code-block:: idris
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||| Perform substitution in a term according to some equality.
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replace : forall x, y, prop . (0 rule : x = y) -> prop x -> prop y
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replace Refl prf = prf
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If we supply an equality (x=y) and a proof of a property of x (``prop x``) then we get
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a proof of a property of y (``prop y``).
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So, in the following example, if we supply ``p1 x`` which is a proof that ``x=2`` and
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the equality ``x=y`` then we get a proof that ``y=2``.
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.. code-block:: idris
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p1: Nat -> Type
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p1 n = (n=2)
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testReplace: (x=y) -> (p1 x) -> (p1 y)
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testReplace a b = replace a b
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Rewrite
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=======
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In practice, ``replace`` can be
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a little tricky to use because in general the implicit argument ``prop``
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can be hard to infer for the machine, so Idris provides a high level
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syntax which calculates the property and applies ``replace``.
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Example: again we supply ``p1 x`` which is a proof that ``x=2`` and the equality
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``y=x`` then we get a proof that ``y=2``.
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.. code-block:: idris
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p1: Nat -> Type
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p1 x = (x=2)
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testRewrite: (y=x) -> (p1 x) -> (p1 y)
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testRewrite a b = rewrite a in b
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We can think of ``rewrite`` as working in this way:
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* Start with a equation ``x=y`` and a property ``prop : x -> Type``
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* Search for ``x`` in ``prop``
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* Replaces all occurrences of ``x`` with ``y`` in ``prop``.
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That is, we are doing a substitution.
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Notice that here we need to supply reverse equality, i.e. ``y=x`` instead of ``x=y``.
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This is because ``rewrite`` performs the substitution of left part of equality to the right part
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and this substitution is done in the *return type*.
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Thus, here in the return type ``y=2`` we need to apply ``y=x`` in order to match the type of the argument ``x=2``.
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Symmetry and Transitivity
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=========================
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In addition to 'reflexivity' equality also obeys 'symmetry' and 'transitivity'
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and these are also included in the prelude:
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.. code-block:: idris
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||| Symmetry of propositional equality
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sym : forall x, y . (0 rule : x = y) -> y = x
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sym Refl = Refl
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||| Transitivity of propositional equality
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trans : forall a, b, c . (0 l : a = b) -> (0 r : b = c) -> a = c
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trans Refl Refl = Refl
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Heterogeneous Equality
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======================
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Also included in the prelude:
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.. code-block:: idris
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||| Explicit heterogeneous ("John Major") equality. Use this when Idris
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||| incorrectly chooses homogeneous equality for `(=)`.
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||| @ a the type of the left side
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||| @ b the type of the right side
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||| @ x the left side
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||| @ y the right side
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(~=~) : (x : a) -> (y : b) -> Type
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(~=~) x y = (x = y)
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