Instead of having m_interrupted flags in several components. We use a thread_local global variable.
The new approach is much simpler to get right since there is no risk of "forgetting" to propagate
the set_interrupt method to sub-components.
The plan is to support set_interrupt methods and m_interrupted flags only in tactic objects.
We need to support them in tactics and tacticals because we want to implement combinators/tacticals such as (try_for T M) that fails if tactic T does not finish in M ms.
For example, consider the tactic:
try-for (T1 ORELSE T2) 5
It tries the tactic (T1 ORELSE T2) for 5ms.
Thus, if T1 does not finish after 5ms an interrupt request is sent, and T1 is interrupted.
Now, if you do not have a m_interrupted flag marking each tactic, the ORELSE combinator will try T2.
The set_interrupt method for ORELSE tactical should turn on the m_interrupted flag.
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>
