mirror of
https://github.com/idris-lang/Idris2.git
synced 2024-12-17 16:21:46 +03:00
211 lines
6.6 KiB
ReStructuredText
211 lines
6.6 KiB
ReStructuredText
Before we discuss the details of theorem proving in Idris, we will describe
|
|
some fundamental concepts:
|
|
|
|
- Propositions and judgments
|
|
- Boolean and constructive logic
|
|
- Curry-Howard correspondence
|
|
- Definitional and propositional equalities
|
|
- Axiomatic and constructive approaches
|
|
|
|
Propositions and Judgments
|
|
==========================
|
|
|
|
Propositions are the subject of our proofs. Before the proof, we can't
|
|
formally say if they are true or not. If the proof is successful then the
|
|
result is a 'judgment'. For instance, if the ``proposition`` is,
|
|
|
|
+-------+
|
|
| 1+1=2 |
|
|
+-------+
|
|
|
|
When we prove it, the ``judgment`` is,
|
|
|
|
+------------+
|
|
| 1+1=2 true |
|
|
+------------+
|
|
|
|
Or if the ``proposition`` is,
|
|
|
|
+-------+
|
|
| 1+1=3 |
|
|
+-------+
|
|
|
|
we can't prove it is true, but it is still a valid proposition and perhaps we
|
|
can prove it is false so the ``judgment`` is,
|
|
|
|
+-------------+
|
|
| 1+1=3 false |
|
|
+-------------+
|
|
|
|
This may seem a bit pedantic but it is important to be careful: in mathematics
|
|
not every proposition is true or false. For instance, a proposition may be
|
|
unproven or even unprovable.
|
|
|
|
So the logic here is different from the logic that comes from boolean algebra.
|
|
In that case what is not true is false and what is not false is true. The logic
|
|
we are using here does not have this law, the "Law of Excluded Middle", so we
|
|
cannot use it.
|
|
|
|
A false proposition is taken to be a contradiction and if we have a
|
|
contradiction then we can prove anything, so we need to avoid this. Some
|
|
languages, used in proof assistants, prevent contradictions.
|
|
|
|
The logic we are using is called constructive (or sometimes intuitional)
|
|
because we are constructing a 'database' of judgments.
|
|
|
|
Curry-Howard correspondence
|
|
---------------------------
|
|
|
|
So how do we relate these proofs to Idris programs? It turns out that there is
|
|
a correspondence between constructive logic and type theory. They have the same
|
|
structure and we can switch back and forth between the two notations.
|
|
|
|
The way that this works is that a proposition is a type so...
|
|
|
|
.. code-block:: idris
|
|
|
|
Main> 1 + 1 = 2
|
|
2 = 2
|
|
|
|
Main> :t 1 + 1 = 2
|
|
(fromInteger 1 + fromInteger 1) === fromInteger 2 : Type
|
|
|
|
...is a proposition and it is also a type. The
|
|
following will also produce an equality type:
|
|
|
|
|
|
.. code-block:: idris
|
|
|
|
Main> 1 + 1 = 3
|
|
2 = 3
|
|
|
|
Both of these are valid propositions so both are valid equality types. But how
|
|
do we represent a true judgment? That is, how do we denote 1+1=2 is true but not
|
|
1+1=3? A type that is true is inhabited, that is, it can be constructed. An
|
|
equality type has only one constructor 'Refl' so a proof of 1+1=2 is
|
|
|
|
.. code-block:: idris
|
|
|
|
onePlusOne : 1+1=2
|
|
onePlusOne = Refl
|
|
|
|
Now that we can represent propositions as types other aspects of
|
|
propositional logic can also be translated to types as follows:
|
|
|
|
+----------+-------------------+--------------------------+
|
|
| | propositions | example of possible type |
|
|
+----------+-------------------+--------------------------+
|
|
| A | x=y | |
|
|
+----------+-------------------+--------------------------+
|
|
| B | y=z | |
|
|
+----------+-------------------+--------------------------+
|
|
| and | A /\\ B | Pair(x=y,y=z) |
|
|
+----------+-------------------+--------------------------+
|
|
| or | A \\/ B | Either(x=y,y=z) |
|
|
+----------+-------------------+--------------------------+
|
|
| implies | A -> B | (x=y) -> (y=x) |
|
|
+----------+-------------------+--------------------------+
|
|
| for all | y=z | |
|
|
+----------+-------------------+--------------------------+
|
|
| exists | y=z | |
|
|
+----------+-------------------+--------------------------+
|
|
|
|
|
|
And (conjunction)
|
|
-----------------
|
|
|
|
We can have a type which corresponds to conjunction:
|
|
|
|
.. code-block:: idris
|
|
|
|
AndIntro : a -> b -> A a b
|
|
|
|
There is a built in type called 'Pair'.
|
|
|
|
Or (disjunction)
|
|
----------------
|
|
|
|
We can have a type which corresponds to disjunction:
|
|
|
|
.. code-block:: idris
|
|
|
|
data Or : Type -> Type -> Type where
|
|
OrIntroLeft : a -> A a b
|
|
OrIntroRight : b -> A a b
|
|
|
|
There is a built in type called 'Either'.
|
|
|
|
Definitional and Propositional Equalities
|
|
-----------------------------------------
|
|
|
|
We have seen that we can 'prove' a type by finding a way to construct a term.
|
|
In the case of equality types there is only one constructor which is ``Refl``.
|
|
We have also seen that each side of the equation does not have to be identical
|
|
like '2=2'. It is enough that both sides are *definitionally equal* like this:
|
|
|
|
.. code-block:: idris
|
|
|
|
onePlusOne : 1+1=2
|
|
onePlusOne = Refl
|
|
|
|
Both sides of this equation normalise to 2 and so Refl matches and the
|
|
proposition is proved.
|
|
|
|
We don't have to stick to terms; we can also use symbolic parameters so the
|
|
following type checks:
|
|
|
|
.. code-block:: idris
|
|
|
|
varIdentity : m = m
|
|
varIdentity = Refl
|
|
|
|
If a proposition/equality type is not definitionally equal but is still true
|
|
then it is *propositionally equal*. In this case we may still be able to prove
|
|
it but some steps in the proof may require us to add something into the terms
|
|
or at least to take some sideways steps to get to a proof.
|
|
|
|
Especially when working with equalities containing variable terms (inside
|
|
functions) it can be hard to know which equality types are definitionally equal,
|
|
in this example ``plusReducesL`` is *definitionally equal* but ``plusReducesR`` is
|
|
not (although it is *propositionally equal*). The only difference between
|
|
them is the order of the operands.
|
|
|
|
.. code-block:: idris
|
|
|
|
plusReducesL : (n:Nat) -> plus Z n = n
|
|
plusReducesL n = Refl
|
|
|
|
plusReducesR : (n:Nat) -> plus n Z = n
|
|
plusReducesR n = Refl
|
|
|
|
Checking ``plusReducesR`` gives the following error:
|
|
|
|
|
|
.. code-block:: idris
|
|
|
|
Proofs.idr:21:18--23:1:While processing right hand side of Main.plusReducesR at Proofs.idr:21:1--23:1:
|
|
Can't solve constraint between:
|
|
plus n Z
|
|
and
|
|
n
|
|
|
|
So why is ``Refl`` able to prove some equality types but not others?
|
|
|
|
The first answer is that ``plus`` is defined by recursion on its first
|
|
argument. So, when the first argument is ``Z``, it reduces, but not when the
|
|
second argument is ``Z``.
|
|
|
|
If an equality type can be proved/constructed by using ``Refl`` alone it is known
|
|
as a *definitional equality*. In order to be definitionally equal both sides
|
|
of the equation must normalise to the same value.
|
|
|
|
So when we type ``1+1`` in Idris it is immediately reduced to 2 because
|
|
definitional equality is built in
|
|
|
|
.. code-block:: idris
|
|
|
|
Main> 1+1
|
|
2
|
|
|
|
In the following pages we discuss how to resolve propositional equalities.
|