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,


When we prove it, the judgment is,

1+1=2 true

Or if the proposition is,


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…

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:

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

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:


example of possible type






A /\ B



A \/ B



A -> B

(x=y) -> (y=z)

for all




And (conjunction)

We can have a type which corresponds to conjunction:

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:

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:

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:

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.

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:

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

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

Main> 1+1

In the following pages we discuss how to resolve propositional equalities.