west, opp is the smallest predicate containing the two sub-cases east_west and
weast_east. If you have some familiarity with Prolog, you may look at opp as the
predicate definined by the two clauses - in this case, the two facts -
-(opp east west) and (opp west east).
-At the end of this section we shall prove that forall a and b,
- (opp a b) iff a = opposite b.
-For the moment, it is time to proceed with our formalization of the farmer's
-problem.
+
+Between opp and opposite we have the following relation:
+ opp a b iff a = opposite b
+Let us prove it, starting from the left to right implication, first *)
+
+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.
+
+(*
+It is time to proceed with our formalization of the farmer's problem.
A state of the system is defined by the position of four item: the goat, the
wolf, the cabbage, and the boat. The simplest way to declare such a data type
is to use a record.
need a few more applications to handle reachability, and side conditions.
The magic number to let automation work is, in this case, 9. *)
-(* lemma problem: \ 5a href="cic:/matita/tutorial/chapter1/reachable.ind(1,0,0)"\ 6reachable\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/start.def(1)"\ 6start\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/end.def(1)"\ 6end\ 5/a\ 6.
-normalize /9/ qed. *)
+lemma problem: \ 5a href="cic:/matita/tutorial/chapter1/reachable.ind(1,0,0)"\ 6reachable\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/start.def(1)"\ 6start\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/end.def(1)"\ 6end\ 5/a\ 6.
+normalize /9/ qed.
(* Let us now try to derive the proof in a more interactive way. Of course, we
expect to need several moves to transfer all items from a bank to the other, so
we should start our proof by applying "more".
*)
-(*
lemma problem1: \ 5a href="cic:/matita/tutorial/chapter1/reachable.ind(1,0,0)"\ 6reachable\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/start.def(1)"\ 6start\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/end.def(1)"\ 6end\ 5/a\ 6.
normalize @\ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"\ 6more\ 5/a\ 6
(* Let us perform the next step, namely moving back the boat, in a sligtly
different way. The more operation expects as second argument the new
-intermediate state, hence instead of applying more we ca
-
-(* We start the proof introducing a, b and the hypothesis opp a b, that we
-call oppab. *)
-#a #b #oppab
+intermediate state, hence instead of applying more we can apply this term
+already instatated on the next intermediate state. As first argument, we
+type a question mark that stands for an implicit argument to be guessed by
+the system. *)
-(* 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 *)
+@(\ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"\ 6more\ 5/a\ 6 ? (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,1,0)"\ 6east\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"\ 6west\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"\ 6west\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"\ 6west\ 5/a\ 6))
-cases oppab // qed.
+(* We now get three independent subgoals, all actives, and two of them are
+trivial. We\ 5span style="font-family: Verdana,sans-serif;"\ 6 \ 5/span\ 6can just apply automation to all of them, and it will close the two
+trivial goals. *)
-(* Let us come to the opposite direction. *)
+/2/
-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". *)
+(* Let us come to the next step, that consists in moving the wolf. Suppose that
+instead of specifying the next intermediate state, we prefer to specify the next
+move. In the spirit of the previous example, we can do it in the following way
+*)
->eqa
+@(\ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"\ 6more\ 5/a\ 6 … (\ 5a href="cic:/matita/tutorial/chapter1/move.con(0,2,0)"\ 6move_wolf\ 5/a\ 6 … ))
-(* We conclude the proof by cases on b. *)
+(* The dots stand here for an arbitrary number of implicit arguments, to be
+guessed by the system.
+Unfortunately, the previous move is not enough to fully instantiate the new
+intermediate state: a bank B remains unknown. Automation cannot help here,
+since all goals depend from this bank and automation refuses to close some
+subgoals instantiating other subgoals remaining open (the instantiation could
+be arbitrary). The simplest way to proceed is to focus on the bank, that is
+the fourth subgoal, and explicitly instatiate it. Instead of repeatedly using "|",
+we can perform focusing by typing "4:" as described by the following command. *)
-cases b // qed.
+[4: @\ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,1,0)"\ 6east\ 5/a\ 6] /2/
-(* 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/. *)
+(* Alternatively, we can directly instantiate the bank into the move. Let
+us complete the proof in this, very readable way. *)
-lemma problem_bis: \ 5a href="cic:/matita/tutorial/chapter1/reachable.ind(1,0,0)"\ 6reachable\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/start.def(1)"\ 6start\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter1/end.def(1)"\ 6end\ 5/a\ 6.
-normalize /9/
\ No newline at end of file
+@(\ 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.
projects / helm.git / blobdiff