lean2/tests/lean/bad9.lean

set_option pp::implicit true.
set_option pp::colors   false.
variable N : Type.

check
fun (a : N) (f : N -> N) (H : f a = a),
let calc1 : f a = a := substp (fun x : N, f a = _) (refl (f a)) H
in  calc1.
chore(*): cleanup lean builtin symbols, replace :: with _ Signed-off-by: Leonardo de Moura <leonardo@microsoft.com> 2014-01-09 08:33:52 -08:00			`set_option pp::implicit true.`
			`set_option pp::colors false.`
feat(frontends/lean): use lowercase commands, replace 'endscope' and 'endnamespace' with 'end' Signed-off-by: Leonardo de Moura <leonardo@microsoft.com> 2014-01-05 12:05:08 -08:00			`variable N : Type.`
feat(kernel/metavar): is instantiate_metavars, we are also instantiating assigned metavariables that occur in the local context of unassinged ones. This modification improves the effectiveness of the process_metavar_inst procedure in the Lean elaborator. For example, suppose we have the constraint ctx \|- ?M1[inst:0 ?M2] == a If ?M1 and ?M2 are unassigned, then we have to consider the two possible solutions: ?M1 == a or ?M1 == #0 and ?M2 == a On the other hand, if ?M2 is assigned to b, then we can ignore the second case. Signed-off-by: Leonardo de Moura <leonardo@microsoft.com> 2014-01-02 16:08:32 -08:00
feat(frontends/lean): use lowercase commands, replace 'endscope' and 'endnamespace' with 'end' Signed-off-by: Leonardo de Moura <leonardo@microsoft.com> 2014-01-05 12:05:08 -08:00			`check`
refactor(kernel): remove heterogeneous equality Signed-off-by: Leonardo de Moura <leonardo@microsoft.com> 2014-01-16 15:07:51 -08:00			`fun (a : N) (f : N -> N) (H : f a = a),`
			`let calc1 : f a = a := substp (fun x : N, f a = _) (refl (f a)) H`
feat(kernel/metavar): is instantiate_metavars, we are also instantiating assigned metavariables that occur in the local context of unassinged ones. This modification improves the effectiveness of the process_metavar_inst procedure in the Lean elaborator. For example, suppose we have the constraint ctx \|- ?M1[inst:0 ?M2] == a If ?M1 and ?M2 are unassigned, then we have to consider the two possible solutions: ?M1 == a or ?M1 == #0 and ?M2 == a On the other hand, if ?M2 is assigned to b, then we can ignore the second case. Signed-off-by: Leonardo de Moura <leonardo@microsoft.com> 2014-01-02 16:08:32 -08:00			`in calc1.`