Let us start this tutorial with a simple example based on the following well
known problem.
-\ 5h2 class="section"\ 6The goat, the wolf and the cabbage\ 5/h2\ 6\ 5div class="paragraph"\ 6\ 5/div\ 6A farmer need to transfer a goat, a wolf and a cabbage across a river, but there
+\ 5h2 class="section"\ 6The goat, the wolf and the cabbage\ 5/h2\ 6
+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
(* 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:
*)
(* 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. \ 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.
of a relation (a binary predicate) over states. *)
inductive move : \ 5a href="cic:/matita/tutorial/chapter1/state.ind(1,0,0)"\ 6state\ 5/a\ 6 → \ 5a href="cic:/matita/tutorial/chapter1/state.ind(1,0,0)"\ 6state\ 5/a\ 6 → Prop ≝
-| move_goat: ∀g,g1,w,c. \ 5a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"\ 6opp\ 5/a\ 6 g g1 → move (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g w c g) (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g1 w c g1)
+| move_goat: ∀g,g1,w,c. \ 5a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"\ 6opp\ 5/a\ 6 g g1 → move (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g w c g) (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 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. \ 5a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"\ 6opp\ 5/a\ 6 w w1 → move (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g w c w) (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g w1 c w1)
| move_cabbage: ∀g,w,c,c1.\ 5a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"\ 6opp\ 5/a\ 6 c c1 → move (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g w c c) (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g w c1 c1)
| move_boat: ∀g,w,c,b,b1. \ 5a href="cic:/matita/tutorial/chapter1/opp.ind(1,0,0)"\ 6opp\ 5/a\ 6 b b1 → move (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g w c b) (\ 5a href="cic:/matita/tutorial/chapter1/state.con(0,1,0)"\ 6mk_state\ 5/a\ 6 g w c b1).
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.
+normalize /\ 5span class="autotactic"\ 69\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,1,0)"\ 6one\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,2,0)"\ 6more\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/safe_state.con(0,1,0)"\ 6with_boat\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/safe_state.con(0,2,0)"\ 6opposite_side\ 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/move.con(0,2,0)"\ 6move_wolf\ 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/move.con(0,4,0)"\ 6move_boat\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"\ 6east_west\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"\ 6west_east\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/ 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
refocus on the skipped goal, going back to a situation similar to the one we
started with. *)
- | /2/ ]
+ | /\ 5span class="autotactic"\ 62\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/tutorial/chapter1/move.con(0,1,0)"\ 6move_goat\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"\ 6east_west\ 5/a\ 6\ 5/span\ 6\ 5/span\ 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
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. *)
-/2/
+/\ 5span class="autotactic"\ 62\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/tutorial/chapter1/safe_state.con(0,2,0)"\ 6opposite_side\ 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/opp.con(0,1,0)"\ 6east_west\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"\ 6west_east\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/
(* 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
the fourth subgoal, and explicitly instatiate it. Instead of repeatedly using "|",
we can perform focusing by typing "4:" as described by the following command. *)
-[4: @\ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,1,0)"\ 6east\ 5/a\ 6] /2/
+[4: @\ 5a href="cic:/matita/tutorial/chapter1/bank.con(0,1,0)"\ 6east\ 5/a\ 6] /\ 5span class="autotactic"\ 62\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/tutorial/chapter1/safe_state.con(0,1,0)"\ 6with_boat\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"\ 6east_west\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/
(* Alternatively, we can directly instantiate the bank into the move. Let
-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/.
+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/ qed. *)
\ 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 … )) /\ 5span class="autotactic"\ 62\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/tutorial/chapter1/safe_state.con(0,1,0)"\ 6with_boat\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"\ 6west_east\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/
+@(\ 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 … )) /\ 5span class="autotactic"\ 62\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/tutorial/chapter1/safe_state.con(0,2,0)"\ 6opposite_side\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"\ 6east_west\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"\ 6west_east\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/
+@(\ 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 … )) /\ 5span class="autotactic"\ 62\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/tutorial/chapter1/safe_state.con(0,1,0)"\ 6with_boat\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,2,0)"\ 6west_east\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/
+@\ 5a href="cic:/matita/tutorial/chapter1/reachable.con(0,1,0)"\ 6one\ 5/a\ 6 /\ 5span class="autotactic"\ 62\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/tutorial/chapter1/move.con(0,1,0)"\ 6move_goat\ 5/a\ 6, \ 5a href="cic:/matita/tutorial/chapter1/opp.con(0,1,0)"\ 6east_west\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/ qed.
\ No newline at end of file
projects / helm.git / blobdiff