For example, this feature is useful when displaying the integer value 10 with coercions enabled. In this case, we want to display "nat_to_int 10" instead of "10".
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
It was not a good idea to use heterogeneous equality as the default equality in Lean.
It creates the following problems.
- Heterogeneous equality does not propagate constraints in the elaborator.
For example, suppose that l has type (List Int), then the expression
l = nil
will not propagate the type (List Int) to nil.
- It is easy to write false. For example, suppose x has type Real, and the user
writes x = 0. This is equivalent to false, since 0 has type Nat. The elaborator cannot introduce
the coercion since x = 0 is a type correct expression.
Homogeneous equality does not suffer from the problems above.
We keep heterogeneous equality because it is useful for generating proof terms.
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
It is incorrect to apply substitutions during normalization.
The problem is that we do not have support for tracking justifications in the normalizer. So, substitutions were being silently applied during normalization. Thus, the correctness of the conflict resolution in the elaboration was being affected.
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
We need that when we normalize the assignment in a metavariable environment.
That is, we replace metavariable in a substitution with other assignments.
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
We may miss solutions, but the solutions found are much more readable.
For example, without this option, for elaboration problem
Theorem Example4 (a b c d e : N) (H: (a = b ∧ b = e ∧ b = c) ∨ (a = d ∧ d = c)) : (h a c) = (h c a) :=
DisjCases H
(fun H1 : _,
let AeqC := Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1))
in CongrH AeqC (Symm AeqC))
(fun H1 : _,
let AeqC := Trans (Conjunct1 H1) (Conjunct2 H1)
in CongrH AeqC (Symm AeqC))
the elaborator generates
Theorem Example4 (a b c d e : N) (H : a = b ∧ b = e ∧ b = c ∨ a = d ∧ d = c) : (h a c) = (h c a) :=
DisjCases
H
(λ H1 : if
Bool
(if Bool (a = b) (if Bool (if Bool (if Bool (b = e) (if Bool (b = c) ⊥ ⊤) ⊤) ⊥ ⊤) ⊥ ⊤) ⊤)
⊥
⊤,
let AeqC := Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1)) in CongrH AeqC (Symm AeqC))
(λ H1 : if Bool (if Bool (a = d) (if Bool (d = c) ⊥ ⊤) ⊤) ⊥ ⊤,
let AeqC := Trans (Conjunct1 H1) (Conjunct2 H1) in CongrH AeqC (Symm AeqC))
The solution is correct, but it is not very readable. The problem is that the elaborator expands the definitions of \/ and /\.
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
Motivations:
- We have been writing several comments of the form "... trace/justification..." and "this trace object justify ...".
- Avoid confusion with util/trace.h
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
This normalization rule is not really a computational rule.
It is essentially encoding the reflexivity axiom as computation.
It can also be abaused. For example, with this rule,
the following definition is valid:
Theorem Th : a = a := Refl b
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
