2 \ 5h1 class="section"
\ 6Induction and Recursion
\ 5/h1
\ 6
4 include "basics/types.ma".
6 (* Most of the types we have seen so far are enumerated types, composed by a
7 finite set of alternatives, and records, composed by tuples of heteregoneous
8 elements. A more interesting case of type definition is when some of the rules
9 defining its elements are recursive, i.e. they allow the formation of more
10 elements of the type in terms of the already defined ones. The most typical case
11 is provided by the natural numbers, that can be defined as the smallest set
12 generated by a constant 0 and a successor function from natural numbers to natural
15 \ 5img class="anchor" src="icons/tick.png" id="nat"
\ 6inductive nat : Type[0] ≝
19 (* The two terms O and S are called constructors: they define the signature of the
20 type, whose objects are the elements freely generated by means of them. So,
21 examples of natural numbers are O, S O, S (S O), S (S (S O)) and so on.
23 The language of Matita allows the definition of well founded recursive functions
24 over inductive types; in order to guarantee termination of recursion you are only
25 allowed to make recursive calls on structurally smaller arguments than the ones
26 you received in input. Most mathematical functions can be naturally defined in this
27 way. For instance, the sum of two natural numbers can be defined as follows *)
29 \ 5img class="anchor" src="icons/tick.png" id="add"
\ 6let rec add n m ≝
32 | S a ⇒
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"
\ 6S
\ 5/a
\ 6 (add a m)
36 \ 5h2 class="section"
\ 6Elimination
\ 5/h2
\ 6
37 It is worth to observe that the previous algorithm works by recursion over the
38 first argument. This means that, for instance, (add O x) will reduce to x, as
39 expected, but (add x O) is stuck.
40 How can we prove that, for a generic x, (add x O) = x? The mathematical tool to do
41 it is called induction. The induction principle states that, given a property P(n)
42 over natural numbers, if we prove P(0) and prove that, for any m, P(m) implies P(S m),
43 than we can conclude P(n) for any n.
45 The elim tactic, allow you to apply induction in a very simple way. If your goal is
46 P(n), the invocation of
48 will break down your task to prove the two subgoals P(0) and ∀m.P(m) → P(S m).
50 Let us apply it to our case *)
52 \ 5img class="anchor" src="icons/tick.png" id="add_0"
\ 6lemma 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.
55 (* If you stop the computation here, you will see on the right the two subgoals
57 - ∀x. add x 0 = x → add (S x) O = S x
58 After normalization, both goals are trivial.
63 (* In a similar way, it is convenient to state a lemma about the behaviour of
64 add when the second argument is not zero. *)
66 \ 5img class="anchor" src="icons/tick.png" id="add_S"
\ 6lemma 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).
68 (* In the same way as before, we proceed by induction over a. *)
70 #a #b elim a normalize //
73 (* We are now in the position to prove the commutativity of the sum *)
75 \ 5img class="anchor" src="icons/tick.png" id="add_comm"
\ 6theorem 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.
78 (* We have two sub goals:
80 G2: ∀x.(∀b. add x b = add b x) → ∀b. S (add x b) = add b (S x).
81 G1 is just our lemma add_O. For G2, we start introducing x and the induction
82 hypothesis IH; then, the goal is proved by rewriting using add_S and IH.
83 For Matita, the task is trivial and we can simply close the goal with // *)
89 \ 5img class="anchor" src="icons/tick.png" id="bool"
\ 6inductive bool : Type[0] ≝
93 \ 5img class="anchor" src="icons/tick.png" id="nat_of_bool"
\ 6definition nat_of_bool ≝ λb. match b with
94 [ 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
95 | ff ⇒
\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"
\ 6O
\ 5/a
\ 6
98 (* coercion nat_of_bool. ?? *)
100 (* Let us now define the following function: *)
102 \ 5img class="anchor" src="icons/tick.png" id="twice"
\ 6definition twice ≝ λn.
\ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"
\ 6add
\ 5/a
\ 6 n n.
105 \ 5h2 class="section"
\ 6Existential
\ 5/h2
\ 6
106 We are interested to prove that for any natural number n there exists a natural
107 number m that is the integer half of n. This will give us the opportunity to
108 introduce new connectives and quantifiers and, later on, to make some interesting
109 consideration on proofs and computations. *)
111 \ 5img class="anchor" src="icons/tick.png" id="ex_half"
\ 6theorem 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).
114 (* We proceed by induction on n, that breaks down to the following goals:
115 G1: ∃m.O = add O O ∨ O = S (add m m)
116 G2: ∀x.(∃m. x = add m m ∨ x = S (add m m))→ ∃m. S x = add m m ∨ S x = S (add m m)
117 The only way we have to prove an existential goal is by exhibiting the witness,
118 that in the case of first goal is O. We do it by apply the term called ex_intro
119 instantiated by the witness. Then, it is clear that we must follow the left branch
120 of the disjunction. One way to do it is by applying the term or_introl, that is
121 the first constructor of the disjunction. However, remembering the names of
122 constructors can be annyoing: we can invoke the application of the n-th
123 constructor of an inductive type (inferred by the current goal) by typing %n. At
124 this point we are left with the subgoal O = add O O, that is closed by
125 computation. It is worth to observe that invoking automation at depth /3/ would
126 also automatically close G1.
128 [@(
\ 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 //
131 \ 5h2 class="section"
\ 6Destructuration
\ 5/h2
\ 6
132 The case of G2 is more complex. We should start introducing x and the
134 IH: ∃m. x = add m m ∨ x = S (add m m)
135 At this point we should assume the existence of m enjoying the inductive
136 hypothesis. To eliminate the existential from the context we can just use the
137 case tactic. This situation where we introduce something into the context and
138 immediately eliminate it by case analysis is so frequent that Matita provides a
139 convenient shorthand: you can just type a single "*".
140 The star symbol should be reminiscent of an explosion: the idea is that you have
141 a structured hypothesis, and you ask to explode it into its constituents. In the
142 case of the existential, it allows to pass from a goal of the shape
144 to a goal of the shape
148 (* At this point we are left with a new goal with the following shape
149 G3: ∀m. x = add m m ∨ x = S (add m m) → ....
150 We should introduce m, the hypothesis H: x = add m m ∨ x = S (add m m), and
151 then reason by cases on this hypothesis. It is the same situation as before:
152 we explode the disjunctive hypothesis into its possible consituents. In the case
153 of a disjunction, the * tactic allows to pass from a goal of the form
155 to two subgoals of the form
159 (* In the first subgoal, we are under the assumption that x = add m m. The half
160 of (S x) is hence m, and we have to prove the right branch of the disjunction.
161 In the second subgoal, we are under the assumption that x = S (add m m). The halh
162 of (S x) is hence (S m), and have to follow the left branch of the disjunction.
164 [@(
\ 5a href="cic:/matita/basics/logic/ex.con(0,1,2)"
\ 6ex_intro
\ 5/a
\ 6 … m) /
\ 5span class="autotactic"
\ 62
\ 5span class="autotrace"
\ 6 trace
\ 5a href="cic:/matita/basics/logic/Or.con(0,2,2)"
\ 6or_intror
\ 5/a
\ 6\ 5/span
\ 6\ 5/span
\ 6/ | @(
\ 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 /
\ 5span class="autotactic"
\ 62
\ 5span class="autotrace"
\ 6 trace
\ 5a href="cic:/matita/basics/logic/Or.con(0,1,2)"
\ 6or_introl
\ 5/a
\ 6\ 5/span
\ 6\ 5/span
\ 6/
169 \ 5h2 class="section"
\ 6Computing vs. Proving
\ 5/h2
\ 6
170 Instead of proving the existence of a number corresponding to the half of n,
171 we could be interested in computing it. The best way to do it is to define this
172 division operation together with the remainder, that in our case is just a
173 boolean value: tt if the input term is even, and ff if the input term is odd.
174 Since we must return a pair, we could use a suitably defined record type, or
175 simply a product type nat × bool, defined in the basic library. The product type
176 is just a sort of general purpose record, with standard fields fst and snd, called
178 A pair of values n and m is written (pair … m n) or \langle n,m \rangle - visually
181 We first write down the function, and then discuss it.*)
183 \ 5img class="anchor" src="icons/tick.png" id="div2"
\ 6let rec div2 n ≝
185 [ 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\ 5span class="error" title="Parse error: [sym,] expected after [term level 19] (in [term])"
\ 6\ 5/span
\ 6,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6
186 | S a ⇒
\ 5span style="text-decoration: underline;"
\ 6\ 5/span
\ 6
188 match (
\ 5a href="cic:/matita/basics/types/snd.fix(0,2,1)"
\ 6snd
\ 5/a
\ 6\ 5span class="error" title="Parse error: SYMBOL ':' or RPAREN expected after [term] (in [term])"
\ 6\ 5/span
\ 6 … p) with
189 [ 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.fix(0,2,1)"
\ 6fst
\ 5/a
\ 6 … p),
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6
190 | ff ⇒
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〈
\ 5/a
\ 6\ 5a href="cic:/matita/basics/types/fst.fix(0,2,1)"
\ 6fst
\ 5/a
\ 6 … p,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,1,0)"
\ 6tt
\ 5/a
\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6
194 (* The function is computed by recursion over the input n. If n is 0, then the
195 quotient is 0 and the remainder is tt. If n = S a, we start computing the half
196 of a, say 〈q,b〉. Then we have two cases according to the possible values of b:
197 if b is tt, then we must return 〈q,ff〉, while if b = ff then we must return
200 It is important to point out the deep, substantial analogy between the algorithm
201 for computing div2 and the the proof of ex_half. In particular ex_half returns a
202 proof of the kind ∃n.A(n)∨B(n): the really informative content in it is the
203 witness n and a boolean indicating which one between the two conditions A(n) and
204 B(n) is met. This is precisely the quotient-remainder pair returned by div2.
205 In both cases we proceed by recurrence (respectively, induction or recursion) over
206 the input argument n. In case n = 0, we conclude the proof in ex_half by providing
207 the witness O and a proof of A(O); this corresponds to returning the pair 〈O,ff〉 in
208 div2. Similarly, in the inductive case n = S a, we must exploit the inductive
209 hypothesis for a (i.e. the result of the recursive call), distinguishing two subcases
210 according to the the two possibilites A(a) or B(a) (i.e. the two possibile values of
211 the remainder for a). The reader is strongly invited to check all remaining details.
213 Let us now prove that our div2 function has the expected behaviour.
216 \ 5img class="anchor" src="icons/tick.png" id="surjective_pairing"
\ 6lemma 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.fix(0,2,1)"
\ 6fst
\ 5/a
\ 6 … p,
\ 5a title="pair pi2" href="cic:/fakeuri.def(1)"
\ 6\snd
\ 5/a
\ 6\ 5span class="error" title="Parse error: [sym〉] or [sym,] expected after [term level 19] (in [term])"
\ 6\ 5/span
\ 6 … p
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6.
219 \ 5img class="anchor" src="icons/tick.png" id="div2SO"
\ 6lemma 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 title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 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\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6.
220 #n #q #H normalize >H normalize // qed.
222 \ 5img class="anchor" src="icons/tick.png" id="div2S1"
\ 6lemma 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 title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 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\ 5span class="error" title="Parse error: [term] expected after [sym=] (in [term])"
\ 6\ 5/span
\ 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\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6.
223 #n #q #H normalize >H normalize // qed.
225 \ 5img class="anchor" src="icons/tick.png" id="div2_ok"
\ 6lemma 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
\ 5a title="Pair construction" 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/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).
227 [#q #r normalize #H destruct //
229 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.fix(0,2,1)"
\ 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.fix(0,2,1)"
\ 6snd
\ 5/a
\ 6 … (
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 a)
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6) [//]
230 cases (
\ 5a href="cic:/matita/basics/types/snd.fix(0,2,1)"
\ 6snd
\ 5/a
\ 6 … (
\ 5a href="cic:/matita/tutorial/chapter2/div2.fix(0,0,2)"
\ 6div2
\ 5/a
\ 6 a))
231 [#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 whd in ⊢ (???%); <
\ 5a href="cic:/matita/tutorial/chapter2/add_S.def(2)"
\ 6add_S
\ 5/a
\ 6 @(Hind … H)
232 |#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)
237 \ 5h2 class="section"
\ 6Mixing proofs and computations
\ 5/h2
\ 6
238 There is still another possibility, however, namely to mix the program and its
239 specification into a single entity. The idea is to refine the output type of the
240 div2 function: it should not be just a generic pair 〈q,r〉 of natural numbers but a
241 specific pair satisfying the specification of the function. In other words, we need
242 the possibility to define, for a type A and a property P over A, the subset type
243 {a:A|P(a)} of all elements a of type A that satisfy the property P. Subset types
244 are just a particular case of the so called dependent types, that is types that
245 can depend over arguments (such as arrays of a specified length, taken as a
246 parameter).These kind of types are quite unusual in traditional programming
247 languages, and their study is one of the new frontiers of the current research on
250 There is nothing special in a subset type {a:A|P(a)}: it is just a record composed
251 by an element of a of type A and a proof of P(a). The crucial point is to have a
252 language reach enough to comprise proofs among its expressions.
255 \ 5img class="anchor" src="icons/tick.png" id="Sub"
\ 6record Sub (A:Type[0]) (P:A → Prop) : Type[0] ≝
259 \ 5img class="anchor" src="icons/tick.png" id="qr_spec"
\ 6definition 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
\ 5a title="Pair construction" 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/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).
261 (* We can now construct a function from n to {p|qr_spec n p} by composing the objects
264 \ 5img class="anchor" src="icons/tick.png" id="div2P"
\ 6definition div2P: ∀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\ 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.
265 \ 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).
267 (* But we can also try do directly build such an object *)
269 \ 5img class="anchor" src="icons/tick.png" id="div2Pagain"
\ 6definition 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).
271 [@(
\ 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\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6) normalize #q #r #H destruct //
273 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.fix(0,2,1)"
\ 6fst
\ 5/a
\ 6 … p,
\ 5a href="cic:/matita/basics/types/snd.fix(0,2,1)"
\ 6snd
\ 5/a
\ 6 … p
\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6) [//]
274 cases (
\ 5a href="cic:/matita/basics/types/snd.fix(0,2,1)"
\ 6snd
\ 5/a
\ 6 … p)
275 [#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.fix(0,2,1)"
\ 6fst
\ 5/a
\ 6 … p),
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,2,0)"
\ 6ff
\ 5/a
\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 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
276 whd in ⊢ (???%); <
\ 5a href="cic:/matita/tutorial/chapter2/add_S.def(2)"
\ 6add_S
\ 5/a
\ 6 @(qrspec … H)
277 |#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.fix(0,2,1)"
\ 6fst
\ 5/a
\ 6 … p,
\ 5a href="cic:/matita/tutorial/chapter2/bool.con(0,1,0)"
\ 6tt
\ 5/a
\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 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)
281 \ 5img class="anchor" src="icons/tick.png" id="quotient7"
\ 6example 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\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6.
284 \ 5img class="anchor" src="icons/tick.png" id="quotient8"
\ 6example 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))))))
285 \ 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\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"
\ 6〉
\ 5/a
\ 6.
287 \ 5pre
\ 6\ 5pre
\ 6 \ 5/pre
\ 6\ 5/pre
\ 6