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It used to be an inline pattern match.

This also restores compatibility with Coq 8.6.1.

saved predicates: use ofe_fun, not ofe_mor

See merge request FP/iris-coq!86

generalize savedProp to let the user control the position of the type-level later

See merge request FP/iris-coq!76

It does not really help since the main work of the proof is in showing that
`cFunctor_map F (iProp_fold, iProp_unfold)` is injective, but whatever.

Fix typo in definition of weakest pre in documentation.

See merge request FP/iris-coq!84

This is an old flag set by the ssr plugin, and recently unset in coq-stdpp,
see https://gitlab.mpi-sws.org/robbertkrebbers/coq-stdpp/issues/5.

Improved treatment of anonymous hypotheses in the proof mode

See merge request FP/iris-coq!83

The proof mode now explicitly keeps track of anonymous hypotheses (i.e.
hypotheses that are introduced by the introduction pattern `?`). Consider:

  Lemma foo {M} (P Q R : uPred M) : P -∗ (Q ∗ R) -∗ Q ∗ P.
  Proof. iIntros "? [H ?]". iFrame "H". iFrame. Qed.

After the `iIntros`, the goal will be:

  _ : P
  "H" : Q
  _ : R
  --------------------------------------∗
  Q ∗ P

Anonymous hypotheses are displayed in a special way (`_ : P`). An important
property of the new anonymous hypotheses is that it is no longer possible to
refer to them by name, whereas before, anonymous hypotheses were given some
arbitrary fresh name (typically prefixed by `~`).

Note tactics can still operate on these anonymous hypotheses. For example, both
`iFrame` and `iAssumption`, as well as the symbolic execution tactics, will
use them. The only thing that is not possible is to refer to them yourself,
for example, in an introduction, specialization or selection pattern.

Advantages of the new approach:

- Proofs become more robust as one cannot accidentally refer to anonymous
  hypotheses by their fresh name.
- Fresh name generation becomes considerably easier. Since anonymous hypotheses
  are internally represented by natural numbers (of type `N`), we can just fold
  over the hypotheses and take the max plus one. This thus solve issue #101.