I read that implications are functions. But I have a hard time trying to understand the example given in the above mentioned page:
The proof term for an implication P → Q is a function that takes evidence for P as input and produces evidence for Q as its output.
Lemma silly_implication : (1 + 1) = 2 → 0 × 3 = 0. Proof. intros H. reflexivity. Qed.
We can see that the proof term for the above lemma is indeed a function:
Print silly_implication. (* ===> silly_implication = fun _ : 1 + 1 = 2 => eq_refl : 1 + 1 = 2 -> 0 * 3 = 0 *)
Indeed, it's a function. But its type does not look right to me. From my reading, the proof term for P -> Q should be a function with an evidence for Q as output. Then, the output of (1+1) = 2 -> 0*3 = 0 should be an evidence for 0*3 = 0, alone, right?
But the Coq print out above shows that the function image is eq_refl : 1 + 1 = 2 -> 0 * 3 = 0, instead of eq_refl: 0 * 3 = 0. I don't understand why the hypothesis 1 + 1 = 2 should appear in the output. Can anyone help explain what is going on here?
Thanks.
Your understanding is correct until:
But the Coq print out above shows that the function image is ...
I think you misunderstand the Print command. Print shows you the term associated with a definition, along with the type of the definition. It does not show the image/output of a function.
For example, the following prints the definition and type of the value x:
Definition x := 5.
Print x.
> x = 5
> : nat
Similarly, the following prints the definition and type of the function f:
Definition f := fun n => n + 2.
Print f.
> f = fun n : nat => n + 2
> : nat -> nat
If you want to see the function's codomain, you have to apply the function to a value, like so:
Definition fx := f x.
Print fx.
> fx = f x
> : nat
If you want to see the image/output of a function Print won't help you. What you need is Compute. Compute takes a term (e.g. a function application) and reduces it as far as possible:
Compute (f x).
> = 7
> : nat