Most notably:
Give le.refl the attribute [refl]. This simplifies tactic proofs in various places.
Redefine the order of trunc_index, and instantiate it as weak order.
Add more about pointed equivalences.
I addressed two problems. First, the theorem names and notation were all in
the namespace complete_lattice. The problem was that if you opened that
namespace, names (like "sup" and "inf") and notation clashed with global notation
for lattices.
The other problem was that if you defined a lattice using Sup, the Sup you got
was not the Sup you want; it was the Sup-construction from the Inf-construction
from the Sup.
Everything seems good now.
@avigad, @fpvandoorn, @rlewis1988, @dselsam
I changed how transitive instances are named.
The motivation is to avoid a naming collision problem found by Daniel.
Before this commit, we were getting an error on the following file
tests/lean/run/collision_bug.lean.
Now, transitive instances contain the prefix "_trans_".
It makes it clear this is an internal definition and it should not be used
by users.
This change also demonstrates (again) how the `rewrite` tactic is
fragile. The problem is that the matching procedure used by it has
very little support for solving matching constraints that involving type
class instances. Eventually, we will need to reimplement `rewrite`
using the new unification procedure used in blast.
In the meantime, the workaround is to use `krewrite` (as usual).
Before this commit, Lean would forbid new tokens containing '(' or ')'.
We relax this restriction. Now, we just forbid new tokens starting with '(' or ending with ')'.
The motivation is to reduce the number of instances generated by ematching.
For example, given
inv_inv: forall a, (a⁻¹)⁻¹ = a
the new heuristic uses ((a⁻¹)⁻¹) as the pattern.
This matches the intuition that inv_inv should be used a simplification
rule.
The default pattern inference procedure would use (a⁻¹). This is bad
because it generates an infinite chain of instances whenever there is a
term (a⁻¹) in the proof state.
By using (a⁻¹), we get
(a⁻¹)⁻¹ = a
Now that we have (a⁻¹)⁻¹, we can match again and generate
((a⁻¹)⁻¹)⁻¹ = a⁻¹
and so on
