1 include "basics/types.ma".
3 (* Most of the types we have seen so far are enumerated types, composed by a finite set of
4 alternatives, and records, composed by tuples of heteregoneous elements.
5 A more interesting case of type definition is when some of the rules defining its elements are
6 recursive, i.e. they allow the formation of more elements of the type in terms of the already
7 defined ones. The most typical case is provided by the natural numbers, that can be defined as
8 the smallest set generated by a constant 0 and a successor function from natural numbers to
11 inductive nat : Type[0] ≝
15 (* The two terms O and S are called constructors: they define the signature of the type, whose
16 objects are the elements freely generated by means of them. So, examples of natural numbers are
17 O, S O, S (S O), S (S (S O)) and so on.
19 The language of Matita allows the definition of well founded recursive functions over inductive
20 types; in order to guarantee termination of recursion you are only allowed to make recursive
21 calls on structurally smaller arguments than the ones you received in input.
22 Most mathematical functions can be naturally defined in this way. For instance, the sum of two
23 natural numbers can be defined as follows *)
28 | S a ⇒
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 (add a m)
31 (* It is worth to observe that the previous algorithm works by recursion over the first
32 argument. This means that, for instance, (add O x) will reduce to x, as expected, but (add x O)
33 is stuck. How can we prove that, for a generic x, (add x O) = x? The mathematical tool to do it
34 is called induction. The induction principle states that, given a property P(n) over natural
35 numbers, if we prove P(0) and prove that, for any m, P(m) implies P(S m), than we can conclude
38 The elim tactic, allow you to apply induction in a very simple way. If your goal is P(n), the
41 will break down your task to prove the two subgoals P(0) and ∀m.P(m) → P(S m).
43 Let us apply it to our case *)
45 lemma add_0: ∀a.
\ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 a
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6 \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 a.
48 (* If you stop the computation here, you will see on the right the two subgoals
50 - ∀x. add x 0 = x → add (S x) O = S x
51 After normalization, both goals are trivial.
56 (* In a similar way, it is convenient to state a lemma about the behaviour of add when the
57 second argument is not zero. *)
59 lemma add_S : ∀a,b.
\ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 a (
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 b)
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 a b).
61 (* In the same way as before, we proceed by induction over a. *)
63 #a #b elim a normalize //
66 (* We are now in the position to prove the commutativity of the sum *)
68 theorem add_comm : ∀a,b.
\ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 a b
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 b a.
71 (* We have two sub goals:
73 G2: ∀x.(∀b. add x b = add b x) → ∀b. S (add x b) = add b (S x).
74 G1 is just our lemma add_O. For G2, we start introducing x and the induction hypothesis IH;
75 then, the goal is proved by rewriting using add_S and IH.
76 For Matita, the task is trivial and we can simply close the goal with // *)
82 inductive bool : Type[0] ≝
86 definition nat_of_bool ≝ λb. match b with
87 [ tt ⇒
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6
88 | ff ⇒
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6
91 (* coercion nat_of_bool. ?? *)
93 (* Let us now define the following function: *)
95 definition twice ≝ λn.
\ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 n n.
97 (* We are interested to prove that for any natural number n there exists a natural number m that
98 is the integer half of n. This will give us the opportunity to introduce new connectives and
99 quantifiers and, later on, to make some interesting consideration on proofs and computations. *)
101 theorem ex_half: ∀n.
\ 5a title="exists" href="cic:/fakeuri.def(1)"
\ 6∃
\ 5/a
\ 6m. n
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 m
\ 5a title="logical or" href="cic:/fakeuri.def(1)"
\ 6∨
\ 5/a
\ 6 n
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 m).
104 (* We proceed by induction on n, that breaks down to the following goals:
105 G1: ∃m.O = add O O ∨ O = S (add m m)
106 G2: ∀x.(∃m. x = add m m ∨ x = S (add m m))→ ∃m. S x = add m m ∨ S x = S (add m m)
107 The only way we have to prove an existential goal is by exhibiting the witness, that in the case
108 of first goal is O. We do it by apply the term called ex_intro instantiated by the witness.
109 Then, it is clear that we must follow the left branch of the disjunction. One way to do it is by
110 applying the term or_introl, that is the first constructor of the disjunction. However,
111 remembering the names of constructors can be annyoing: we can invoke the application of the n-th
112 constructor of an inductive type (inferred by the current goal) by typing %n. At this point
113 we are left with the subgoal O = add O O, that is closed by computation.
114 It is worth to observe that invoking automation at depth /3/ would also automatically close G1.
116 [@(
\ 5a href="cic:/matita/basics/logic/ex.con(0,1,2)"
\ 6ex_intro
\ 5/a
\ 6 …
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6) %1 //
118 (* The case of G2 is more complex. We should start introducing x and the inductive hypothesis
119 IH: ∃m. x = add m m ∨ x = S (add m m)
120 At this point we should assume the existence of m enjoying the inductive hypothesis. To
121 eliminate the existential from the context we can just use the case tactic. This situation where
122 we introduce something into the context and immediately eliminate it by case analysis is so
123 frequent that Matita provides a convenient shorthand: you can just type a single "*".
124 The star symbol should be reminiscent of an explosion: the idea is that you have a structured
125 hypothesis, and you ask to explode it into its constituents. In the case of the existential,
126 it allows to pass from a goal of the shape
128 to a goal of the shape
132 (* At this point we are left with a new goal with the following shape
133 G3: ∀m. x = add m m ∨ x = S (add m m) → ....
134 We should introduce m, the hypothesis H: x = add m m ∨ x = S (add m m), and then reason by
135 cases on this hypothesis. It is the same situation as before: we explode the disjunctive
136 hypothesis into its possible consituents. In the case of a disjunction, the * tactic allows
137 to pass from a goal of the form
139 to two subgoals of the form
143 (* In the first subgoal, we are under the assumption that x = add m m. The half of (S x)
144 is hence m, and we have to prove the right branch of the disjunction.
145 In the second subgoal, we are under the assumption that x = S (add m m). The halh of (S x)
146 is hence (S m), and have to follow the left branch of the disjunction.
148 [@(
\ 5a href="cic:/matita/basics/logic/ex.con(0,1,2)"
\ 6ex_intro
\ 5/a
\ 6 … m) /2/ | @(
\ 5a href="cic:/matita/basics/logic/ex.con(0,1,2)"
\ 6ex_intro
\ 5/a
\ 6 … (
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 m)) normalize /2/
152 (* Instead of proving the existence of a number corresponding to the half of n, we could
153 be interested in computing it. The best way to do it is to define this division operation
154 together with the remainder, that in our case is just a boolean value: tt if the input term
155 is even, and ff if the input term is odd. Since we must return a pair, we could use a
156 suitably defined record type, or simply a product type nat × bool, defined in the basic library.
157 The product type is just a sort of general purpose record, with standard fields fst and snd,
158 called projections. A pair of values n and m is written (pair … m n) or \langle n,m \rangle -
159 visually rendered as 〈n,m〉
161 We first write down the function, and then discuss it.*)
165 [ O ⇒
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6〉
166 | S a ⇒
\ 5span style="text-decoration: underline;"
\ 6\ 5/span
\ 6
168 match (
\ 5a href="cic:/matita/basics/types/snd.def(1)"
\ 6snd
\ 5/a
\ 6 … p) with
169 [ tt ⇒
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 (
\ 5a href="cic:/matita/basics/types/fst.def(1)"
\ 6fst
\ 5/a
\ 6 … p),
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6〉
170 | ff ⇒
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/basics/types/fst.def(1)"
\ 6fst
\ 5/a
\ 6 … p,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,1,0)"
\ 6tt
\ 5/a
\ 6〉
174 (* The function is computed by recursion over the input n. If n is 0, then the quotient
175 is 0 and the remainder is tt. If n = S a, we start computing the half of a, say 〈q,b〉.
176 Then we have two cases according to the possible values of b: if b is tt, then we must return
177 〈q,ff〉, while if b = ff then we must return 〈S q,tt〉.
179 It is important to point out the deep, substantial analogy between the algorithm for
180 computing div2 and the the proof of ex_half. In particular ex_half returns a
181 proof of the kind ∃n.A(n)∨B(n): the really informative content in it is the witness
182 n and a boolean indicating which one between the two conditions A(n) and B(n) is met.
183 This is precisely the quotient-remainder pair returned by div2.
184 In both cases we proceed by recurrence (respectively, induction or recursion) over the
185 input argument n. In case n = 0, we conclude the proof in ex_half by providing the
186 witness O and a proof of A(O); this corresponds to returning the pair 〈O,ff〉 in div2.
187 Similarly, in the inductive case n = S a, we must exploit the inductive hypothesis
188 for a (i.e. the result of the recursive call), distinguishing two subcases according
189 to the the two possibilites A(a) or B(a) (i.e. the two possibile values of the remainder
190 for a). The reader is strongly invited to check all remaining details.
192 Let us now prove that our div2 function has the expected behaviour.
195 lemma surjective_pairing: ∀A,B.∀p:A
\ 5a title="Product" href="cic:/fakeuri.def(1)"
\ 6×
\ 5/a
\ 6B. p
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/basics/types/fst.def(1)"
\ 6fst
\ 5/a
\ 6 … p,
\ 5a title="pair pi2" href="cic:/fakeuri.def(1)"
\ 6\snd
\ 5/a
\ 6 … p〉.
198 lemma div2SO: ∀n,q.
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 n
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6q,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6〉 →
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 n)
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5span style="text-decoration: underline;"
\ 6\ 5/span
\ 6q,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,1,0)"
\ 6tt
\ 5/a
\ 6〉.
199 #n #q #H normalize >H normalize // qed.
201 lemma div2S1: ∀n,q.
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 n
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6q,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,1,0)"
\ 6tt
\ 5/a
\ 6〉 →
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 n)
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 q,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6〉.
202 #n #q #H normalize >H normalize // qed.
204 lemma div2_ok: ∀n,q,r.
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 n
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6q,r〉 → n
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 q) (
\ 5a href="cic:/matita/tutorial/chapter2/nat_of_bool.def(1)"
\ 6nat_of_bool
\ 5/a
\ 6 r).
206 [#q #r normalize #H destruct //
208 cut (
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 a
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/basics/types/fst.def(1)"
\ 6fst
\ 5/a
\ 6 … (
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 a),
\ 5a href="cic:/matita/basics/types/snd.def(1)"
\ 6snd
\ 5/a
\ 6 … (
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 a)〉) [//]
209 cases (
\ 5a href="cic:/matita/basics/types/snd.def(1)"
\ 6snd
\ 5/a
\ 6 … (
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 a))
210 [#H >(
\ 5a href="cic:/matita/tutorial/chapter2/div2S1.def(3)"
\ 6div2S1
\ 5/a
\ 6 … H) #H1 destruct @
\ 5a href="cic:/matita/basics/logic/eq_f.def(3)"
\ 6eq_f
\ 5/a
\ 6 \ 5span style="text-decoration: underline;"
\ 6>
\ 5/span
\ 6\ 5a href="cic:/matita/tutorial/chapter2/add_S.def(2)"
\ 6add_S
\ 5/a
\ 6 <
\ 5a href="cic:/matita/tutorial/chapter2/add_S.def(2)"
\ 6add_S
\ 5/a
\ 6 @(Hind … H)
211 |#H >(
\ 5a href="cic:/matita/tutorial/chapter2/div2SO.def(3)"
\ 6div2SO
\ 5/a
\ 6 … H) #H1 destruct >
\ 5a href="cic:/matita/tutorial/chapter2/add_S.def(2)"
\ 6add_S
\ 5/a
\ 6 @
\ 5a href="cic:/matita/basics/logic/eq_f.def(3)"
\ 6eq_f
\ 5/a
\ 6 @(Hind … H)
215 (* There is still another possibility, however, namely to mix the program and its
216 specification into a single entity. The idea is to refine the output type of the div2
217 function: it should not be just a generic pair 〈q,r〉 of natural numbers but a specific
218 pair satisfying the specification of the function. In other words, we need the
219 possibility to define, for a type A and a property P over A, the subset type
220 {a:A|P(a)} of all elements a of type A that satisfy the property P. Subset types are
221 just a particular case of the so called dependent types, that is types that can
222 depend over arguments (such as arrays of a specified length, taken as a parameter).
223 These kind of types are quite unusual in traditional programming languages, and their
224 study is one of the new frontiers of the current research on type systems.
226 There is nothing special in a subset type {a:A|P(a)}: it is just a record composed by
227 an element of a of type A and a proof of P(a). The crucial point is to have a language
228 reach enough to comprise proofs among its expressions.
231 record Sub (A:Type[0]) (P:A → Prop) : Type[0] ≝
235 definition qr_spec ≝ λn.λp.∀q,r. p
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6q,r〉 → n
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 q) (
\ 5a href="cic:/matita/tutorial/chapter2/nat_of_bool.def(1)"
\ 6nat_of_bool
\ 5/a
\ 6 r).
237 (* We can now construct a function from n to {p|qr_spec n p} by composing the objects
240 definition div2P: ∀n.
\ 5a href="cic:/matita/tutorial/chapter2/Sub.ind(1,0,2)"
\ 6 Sub
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/nat.ind(1,0,0)"
\ 6nat
\ 5/a
\ 6\ 5a title="Product" href="cic:/fakeuri.def(1)"
\ 6×
\ 5/a
\ 6\ 5span style="text-decoration: underline;"
\ 6\ 5a href="cic:/matita/tutorial/chapter2/bool.ind(1,0,0)"
\ 6bool
\ 5/a
\ 6\ 5/span
\ 6) (
\ 5a href="cic:/matita/tutorial/chapter2/qr_spec.def(3)"
\ 6qr_spec
\ 5/a
\ 6 n) ≝ λn.
241 \ 5a href="cic:/matita/tutorial/chapter2/Sub.con(0,1,2)"
\ 6mk_Sub
\ 5/a
\ 6 ?? (
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 n) (
\ 5a href="cic:/matita/tutorial/chapter2/div2_ok.def(4)"
\ 6div2_ok
\ 5/a
\ 6 n).
243 (* But we can also try do directly build such an object *)
245 definition div2Pagain : ∀n.
\ 5a href="cic:/matita/tutorial/chapter2/Sub.ind(1,0,2)"
\ 6Sub
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/nat.ind(1,0,0)"
\ 6nat
\ 5/a
\ 6\ 5a title="Product" href="cic:/fakeuri.def(1)"
\ 6×
\ 5/a
\ 6\ 5span style="text-decoration: underline;"
\ 6\ 5/span
\ 6\ 5a href="cic:/matita/tutorial/chapter2/bool.ind(1,0,0)"
\ 6bool
\ 5/a
\ 6) (
\ 5a href="cic:/matita/tutorial/chapter2/qr_spec.def(3)"
\ 6qr_spec
\ 5/a
\ 6 n).
247 [@(
\ 5a href="cic:/matita/tutorial/chapter2/Sub.con(0,1,2)"
\ 6mk_Sub
\ 5/a
\ 6 …
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6〉) normalize #q #r #H destruct //
249 cut (p
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/basics/types/fst.def(1)"
\ 6fst
\ 5/a
\ 6 … p,
\ 5a href="cic:/matita/basics/types/snd.def(1)"
\ 6snd
\ 5/a
\ 6 … p〉) [//]
250 cases (
\ 5a href="cic:/matita/basics/types/snd.def(1)"
\ 6snd
\ 5/a
\ 6 … p)
251 [#H @(
\ 5a href="cic:/matita/tutorial/chapter2/Sub.con(0,1,2)"
\ 6mk_Sub
\ 5/a
\ 6 …
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 (
\ 5a href="cic:/matita/basics/types/fst.def(1)"
\ 6fst
\ 5/a
\ 6 … p),
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6〉) whd #q #r #H1 destruct @
\ 5a href="cic:/matita/basics/logic/eq_f.def(3)"
\ 6eq_f
\ 5/a
\ 6 \ 5span style="text-decoration: underline;"
\ 6>
\ 5/span
\ 6\ 5a href="cic:/matita/tutorial/chapter2/add_S.def(2)"
\ 6add_S
\ 5/a
\ 6 <
\ 5a href="cic:/matita/tutorial/chapter2/add_S.def(2)"
\ 6add_S
\ 5/a
\ 6 @(qrspec … H)
252 |#H @(
\ 5a href="cic:/matita/tutorial/chapter2/Sub.con(0,1,2)"
\ 6mk_Sub
\ 5/a
\ 6 …
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/basics/types/fst.def(1)"
\ 6fst
\ 5/a
\ 6 … p,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,1,0)"
\ 6tt
\ 5/a
\ 6〉) whd #q #r #H1 destruct >
\ 5a href="cic:/matita/tutorial/chapter2/add_S.def(2)"
\ 6add_S
\ 5/a
\ 6 @
\ 5a href="cic:/matita/basics/logic/eq_f.def(3)"
\ 6eq_f
\ 5/a
\ 6 @(qrspec … H)
256 example quotient7:
\ 5a href="cic:/matita/tutorial/chapter2/witness.fix(0,2,1)"
\ 6witness
\ 5/a
\ 6 … (
\ 5a href="cic:/matita/tutorial/chapter2/div2Pagain.def(4)"
\ 6div2Pagain
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6(
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6(
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6(
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6(
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6(
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6(
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6))))))))
\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6(
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6(
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6)),
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,1,0)"
\ 6tt
\ 5/a
\ 6〉.
259 example quotient8:
\ 5a href="cic:/matita/tutorial/chapter2/witness.fix(0,2,1)"
\ 6witness
\ 5/a
\ 6 … (
\ 5a href="cic:/matita/tutorial/chapter2/div2Pagain.def(4)"
\ 6div2Pagain
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6))))))
260 \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"
\ 6=
\ 5/a
\ 6 \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"
\ 6twice
\ 5/a
\ 6 (
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6))),
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6〉.
262 \ 5pre
\ 6\ 5pre
\ 6 \ 5/pre
\ 6\ 5/pre
\ 6