Idris2/docs/source/proofs/propositional.rst

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