-us complete the proof in this way. *)
-
-@(\ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"\ 6more\ 5/a\ 6 … (\ 5a href="cic:/matita/tutorial/chapter1/move.con(0,1,0)"\ 6move_goat\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"\ 6west\ 5/a\ 6 … )) /2/
-@(\ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"\ 6more\ 5/a\ 6 … (\ 5a href="cic:/matita/tutorial/chapter1/move.con(0,3,0)"\ 6move_cabbage\ 5/a\ 6 ?? \ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,1,0)"\ 6east\ 5/a\ 6 … )) /2/
-@(\ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"\ 6more\ 5/a\ 6 … (\ 5a href="cic:/matita/tutorial/chapter1/move.con(0,4,0)"\ 6move_boat\ 5/a\ 6 ??? \ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"\ 6west\ 5/a\ 6 … )) /2/
-@\ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,1,0)"\ 6one\ 5/a\ 6 /2/ qed.
-
-(* Let us now go back to the problem of proving that, for all a and b,
- (opp a b) iff a = opposite b.
-Let us start from the first implication. *)
-
-lemma opp_to_opposite: ∀a,b. \ 5a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"\ 6opp\ 5/a\ 6 a b → a \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/opposite.def(1)"\ 6opposite\ 5/a\ 6 b.
-
-(* We start the proof introducing a, b and the hypothesis opp a b, that we
-call oppab. *)
-#a #b #oppab
-
-(* Now we proceed by cases on the possible proofs of (opp a b), that is on the
-possible shapes of oppab. By definition, there are only two possibilities,
-namely east_west or west_east. Both subcases are trivial, and can be closed by
-automation *)
-
-cases oppab // qed.
-
-(* Let us come to the opposite direction. *)
-
-lemma opposite_to_opp: ∀a,b. a \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/opposite.def(1)"\ 6opposite\ 5/a\ 6 b → \ 5a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"\ 6opp\ 5/a\ 6 a b.
-
-(* As usual, we start introducing a, b and the hypothesis (a = opposite b),
-that we call eqa. *)
-
-#a #b #eqa
-
-(* The right way to proceed, now, is by rewriting a into (opposite b). We do
-this by typing ">eqa". If we wished to rewrite in the opposite direction, namely
-opposite b into a, we would have typed "<eqa". *)
-
->eqa
-
-(* We conclude the proof by cases on b. *)
-
-cases b // qed.
-
-(* Let's do now an important remark.
-Comment both problem and problem1, state another time the problem of the goat, the
-wolf and the cabbage, and try again to run automation at level /9/. *)