The following snippet compiles fine in Idris 2,
divMod2 : Nat -> (Nat, Bool)
divMod2 Z = (Z, False)
divMod2 (S Z) = (Z, True)
divMod2 (S (S n)) = case divMod2 n of (k, r) => (S k, r)
lm : {q:_} -> divMod2 (q+q) = (q, False)
lm {q=0} = Refl
lm {q=S q} = rewrite sym $ plusSuccRightSucc q q in
rewrite lm {q} in Refl
but it caused Idris 1.3.4 to fail:
|
14 | rewrite lm {q} in Refl
| ~~~~~~
When checking right hand side of lm with expected type
divMod2 (S q + S q) = (S q, False)
When checking an application of function rewrite__impl:
Type mismatch between
(q1, False) = (q1, False) (Type of Refl)
and
case divMod2 (plus q q) of (k, r) => (S k, r) = (q1, False) (Expected type)
Specifically:
Type mismatch between
(q1, False)
and
case divMod2 (plus q q) of
(k, r) => (S k, r)
I know Idris 1 has become obsolete, but am still curious about the underlying reason behind this observation. Could someone help to explain this?
I've done a bit more dig here and found surprisingly that Idris1 failed to compile the following even simpler code snippet, which compiles correctly by Idris2:
f : Nat -> (Nat, Bool)
f Z = (Z, False)
f (S n) = case f n of (k, r) => (S k, r)
h2 : {q:_} -> f (S (q+1)) = case f (q+1) of (k, r) => (S k, r)
h2 = Refl
It spit error message:
|
19 | h2 = Refl
| ~~~~
When checking right hand side of h2 with expected type
f (S (q + 1)) = case f (q + 1) of (k, r) => (S k, r)
Type mismatch between
case f (plus q 1) of (k, r) => (S k, r) = case f (plus q 1) of (k, r) => (S k, r) (Type of Refl)
and
case f (plus q 1) of (k, r) => (S k, r) = case f (plus q 1) of (k, r) => (S k, r) (Expected type)
Specifically:
Type mismatch between
case f (plus q 1) of
(k, r) => (S k, r)
and
case f (plus q 1) of
(k, r) => (S k, r)
Aren't the two mismatching terms identical? Why couldn't be constructed with Refl?
