How can I use a comparison of to rational numbers in an if-statement?
if 1 = 2 then 1 else 2
1 = 2
is of course Prop
and not bool
.
I don't understand how dfan's answer is related to the question...
Of course, 1 = 2
is a Prop
, it is the statement that 1 is equal to 2. Hopefully you don't have a proof of this statement...
What you want is a function that, given two natural numbers, 1
and 2
, returns true
if they are equal, and false
if they aren't.
The library Coq.Arith.EqNat
gives you such a function, named beq_nat
.
In fact, you might want something even better, a function that returns a proof of equality or a proof of difference:
(* In Coq.Arith.Peano_dec *)
Theorem eq_nat_dec : forall n m, {n = m} + {n <> m}.
(* ^ a proof that n = m
^ or a proof that n <> m *)
if
is overloaded to work with such things, so you can even write:
if eq_nat_dec 2 3 then ... else ...
Qeq_bool
indeed takes two rationals and produces a bool.
Require Export QArith_base.
Eval compute in Qeq_bool (3#2) (3#2). = true: bool
Eval compute in Qeq_bool (3#2) (5#2). = false: bool
Eval compute in (if Qeq_bool (3#2) (5#2) then 7 else 9). = 9: nat
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