X-Git-Url: http://matita.cs.unibo.it/gitweb/?a=blobdiff_plain;f=weblib%2Ftutorial%2Fchapter1.ma;h=9043bdad712795d8f6f76104e72d4dc6601a3455;hb=a2ba04cd90e76937720b16e37cb12a39b46181e3;hp=da0ed93fab4462bcd139fbee1b07fe1dac7a825c;hpb=e8539409764826d67bae1df0b4b7f6aeda0c17c7;p=helm.git

diff --git a/weblib/tutorial/chapter1.ma b/weblib/tutorial/chapter1.ma
index da0ed93fa..9043bdad7 100644
--- a/weblib/tutorial/chapter1.ma
+++ b/weblib/tutorial/chapter1.ma
@@ -23,7 +23,8 @@ computational mechanism based on the declaration of inductive types.
 Let us start this tutorial with a simple example based on the following well 
 known problem.
 
-h2 class="section"The goat, the wolf and the cabbage/h2div class="paragraph"/divA farmer need to transfer a goat, a wolf and a cabbage across a river, but there
+h2 class="section"The goat, the wolf and the cabbage/h2
+A farmer need to transfer a goat, a wolf and a cabbage across a river, but there
 is only one place available on his boat. Furthermore, the goat will eat the 
 cabbage if they are left alone on the same bank, and similarly the wolf will eat
 the goat. The problem consists in bringing all three items safely across the 
@@ -137,7 +138,7 @@ them in turn, in a way that will be described at the end of this section.
 (* Instead of working with functions, it is sometimes convenient to work with
 predicates. For instance, instead of defining a function computing the opposite 
 bank, we could declare a predicate stating when two banks are opposite to each 
-other. Only two cases are possible, leading naturally two the following 
+other. Only two cases are possible, leading naturally to the following 
 definition:
 *)
 
@@ -148,12 +149,47 @@ inductive opp : a href="cic:/matita/tutorial/chapter1/bank.ind(1,0,0)"bank/a
 (* In precisely the same way as "bank" is the smallest type containing east and
 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. 
+predicate defined by the two clauses - in this case, the two facts - ast_west and
+west_east.
+
+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. a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"opp/a a b → a a title="leibnitz's equality" href="cic:/fakeuri.def(1)"=/a a href="cic:/matita/tutorial/chapter1/opposite.def(1)"opposite/a 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 a title="leibnitz's equality" href="cic:/fakeuri.def(1)"=/a a href="cic:/matita/tutorial/chapter1/opposite.def(1)"opposite/a b → a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"opp/a 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.
@@ -176,7 +212,9 @@ definition end ≝ a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_s
 of a relation (a binary predicate) over states. *)
 
 inductive move : a href="cic:/matita/tutorial/chapter1/state.ind(1,0,0)"state/a → a href="cic:/matita/tutorial/chapter1/state.ind(1,0,0)"state/a → Prop ≝
-| move_goat: ∀g,g1,w,c. a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"opp/a g g1 → move (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g w c g) (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g1 w c g1) 
+| move_goat: ∀g,g1,w,c. a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"opp/a g g1 → move (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g w c g) (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g1 w c g1)
+  (* We can move the goat from a bank g to the opposite bank g1 if and only if the
+     boat is on the same bank g of the goat and we move the boat along with it. *)
 | move_wolf: ∀g,w,w1,c. a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"opp/a w w1 → move (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g w c w) (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g w1 c w1)
 | move_cabbage: ∀g,w,c,c1.a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"opp/a c c1 → move (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g w c c) (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g w c1 c1)
 | move_boat: ∀g,w,c,b,b1. a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"opp/a b b1 → move (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g w c b) (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a g w c b1).
@@ -204,15 +242,14 @@ applicative nodes). In this case, there is a solution in six moves, and we
 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: a href="cic:/matita/tutorial/chapter1/reachable.ind(1,0,0)"reachable/a a href="cic:/matita/tutorial/chapter1/start.def(1)"start/a a href="cic:/matita/tutorial/chapter1/end.def(1)"end/a.
-normalize /9/ qed. *)
+lemma problem: a href="cic:/matita/tutorial/chapter1/reachable.ind(1,0,0)"reachable/a a href="cic:/matita/tutorial/chapter1/start.def(1)"start/a a href="cic:/matita/tutorial/chapter1/end.def(1)"end/a.
+normalize /span class="autotactic"9span class="autotrace" trace a href="cic:/matita/tutorial/chapter1/reachable.con(0,1,0)"one/a, a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"more/a, a href="cic:/matita/tutorial/chapter1/safe_state.con(0,1,0)"with_boat/a, a href="cic:/matita/tutorial/chapter1/safe_state.con(0,2,0)"opposite_side/a, a href="cic:/matita/tutorial/chapter1/move.con(0,1,0)"move_goat/a, a href="cic:/matita/tutorial/chapter1/move.con(0,2,0)"move_wolf/a, a href="cic:/matita/tutorial/chapter1/move.con(0,3,0)"move_cabbage/a, a href="cic:/matita/tutorial/chapter1/move.con(0,4,0)"move_boat/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"east_west/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"west_east/a/span/span/ 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: a href="cic:/matita/tutorial/chapter1/reachable.ind(1,0,0)"reachable/a a href="cic:/matita/tutorial/chapter1/start.def(1)"start/a a href="cic:/matita/tutorial/chapter1/end.def(1)"end/a.
 normalize @a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"more/a
 
@@ -267,45 +304,46 @@ requires /2/ since move_goat opens an additional subgoal. By applying "]" we
 refocus on the skipped goal, going back to a situation similar to the one we
 started with. *)
 
-  | /2/ ] 
+  | /span class="autotactic"2span class="autotrace" trace a href="cic:/matita/tutorial/chapter1/move.con(0,1,0)"move_goat/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"east_west/a/span/span/ ] 
 
 (* 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 *)
+@(a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"more/a ? (a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"mk_state/a a href="cic:/matita/tutorial/chapter1/bank.con(0,1,0)"east/a a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"west/a a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"west/a a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"west/a))
 
-cases oppab // qed.
+(* We now get three independent subgoals, all actives, and two of them are 
+trivial. Wespan style="font-family: Verdana,sans-serif;" /spancan just apply automation to all of them, and it will close the two
+trivial goals. *)
 
-(* Let us come to the opposite direction. *)
+/span class="autotactic"2span class="autotrace" trace a href="cic:/matita/tutorial/chapter1/safe_state.con(0,2,0)"opposite_side/a, a href="cic:/matita/tutorial/chapter1/move.con(0,4,0)"move_boat/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"east_west/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"west_east/a/span/span/
 
-lemma opposite_to_opp: ∀a,b. a a title="leibnitz's equality" href="cic:/fakeuri.def(1)"=/a a href="cic:/matita/tutorial/chapter1/opposite.def(1)"opposite/a b → a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"opp/a 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
+@(a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"more/a … (a href="cic:/matita/tutorial/chapter1/move.con(0,2,0)"move_wolf/a … ))
 
-(* 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: @a href="cic:/matita/tutorial/chapter1/bank.con(0,1,0)"east/a] /span class="autotactic"2span class="autotrace" trace a href="cic:/matita/tutorial/chapter1/safe_state.con(0,1,0)"with_boat/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"east_west/a/span/span/
 
-(* 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: a href="cic:/matita/tutorial/chapter1/reachable.ind(1,0,0)"reachable/a a href="cic:/matita/tutorial/chapter1/start.def(1)"start/a a href="cic:/matita/tutorial/chapter1/end.def(1)"end/a.
-normalize /9/  
\ No newline at end of file
+@(a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"more/a … (a href="cic:/matita/tutorial/chapter1/move.con(0,1,0)"move_goat/a a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"west/a … )) /span class="autotactic"2span class="autotrace" trace a href="cic:/matita/tutorial/chapter1/safe_state.con(0,1,0)"with_boat/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"west_east/a/span/span/
+@(a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"more/a … (a href="cic:/matita/tutorial/chapter1/move.con(0,3,0)"move_cabbage/a ?? a href="cic:/matita/tutorial/chapter1/bank.con(0,1,0)"east/a … )) /span class="autotactic"2span class="autotrace" trace a href="cic:/matita/tutorial/chapter1/safe_state.con(0,2,0)"opposite_side/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"east_west/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"west_east/a/span/span/
+@(a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"more/a … (a href="cic:/matita/tutorial/chapter1/move.con(0,4,0)"move_boat/a ??? a href="cic:/matita/tutorial/chapter1/bank.con(0,2,0)"west/a … )) /span class="autotactic"2span class="autotrace" trace a href="cic:/matita/tutorial/chapter1/safe_state.con(0,1,0)"with_boat/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"west_east/a/span/span/
+@a href="cic:/matita/tutorial/chapter1/reachable.con(0,1,0)"one/a /span class="autotactic"2span class="autotrace" trace a href="cic:/matita/tutorial/chapter1/move.con(0,1,0)"move_goat/a, a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"east_west/a/span/span/ qed.
\ No newline at end of file