-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathSuma_divisible.lean
113 lines (99 loc) · 2.75 KB
/
Suma_divisible.lean
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
-- Suma_divisible.lean
-- Si a divide a b y a c, entonces divide a b+c.
-- José A. Alonso Jiménez <https://jaalonso.github.io>
-- Sevilla, 7-noviembre-2023
-- ---------------------------------------------------------------------
-- ---------------------------------------------------------------------
-- Demostrar que si a es un divisor de b y de c, tambien lo es de b + c.
-- ----------------------------------------------------------------------
-- Demostración en lenguaje natural
-- ================================
-- Puesto que a divide a b y a c, existen d y e tales que
-- b = ad (1)
-- c = ae (2)
-- Por tanto,
-- b + c = ad + c [por (1)]
-- = ad + ae [por (2)]
-- = a(d + e) [por la distributiva]
-- Por consiguiente, a divide a b + c.
-- Demostraciones con Lean4
-- ========================
import Mathlib.Tactic
variable {a b c : ℕ}
-- 1ª demostración
example
(h1 : a ∣ b)
(h2 : a ∣ c)
: a ∣ (b + c) :=
by
rcases h1 with ⟨d, beq : b = a * d⟩
rcases h2 with ⟨e, ceq: c = a * e⟩
have h1 : b + c = a * (d + e) :=
calc b + c
= (a * d) + c := congrArg (. + c) beq
_ = (a * d) + (a * e) := congrArg ((a * d) + .) ceq
_ = a * (d + e) := by rw [← mul_add]
show a ∣ (b + c)
exact Dvd.intro (d + e) h1.symm
-- 2ª demostración
example
(h1 : a ∣ b)
(h2 : a ∣ c)
: a ∣ (b + c) :=
by
rcases h1 with ⟨d, beq : b = a * d⟩
rcases h2 with ⟨e, ceq: c = a * e⟩
have h1 : b + c = a * (d + e) := by linarith
show a ∣ (b + c)
exact Dvd.intro (d + e) h1.symm
-- 3ª demostración
example
(h1 : a ∣ b)
(h2 : a ∣ c)
: a ∣ (b + c) :=
by
rcases h1 with ⟨d, beq : b = a * d⟩
rcases h2 with ⟨e, ceq: c = a * e⟩
show a ∣ (b + c)
exact Dvd.intro (d + e) (by linarith)
-- 4ª demostración
example
(h1 : a ∣ b)
(h2 : a ∣ c)
: a ∣ (b + c) :=
by
cases' h1 with d beq
-- d : ℕ
-- beq : b = a * d
cases' h2 with e ceq
-- e : ℕ
-- ceq : c = a * e
rw [ceq, beq]
-- ⊢ a ∣ a * d + a * e
use (d + e)
-- ⊢ a * d + a * e = a * (d + e)
ring
-- 5ª demostración
example
(h1 : a ∣ b)
(h2 : a ∣ c)
: a ∣ (b + c) :=
by
rcases h1 with ⟨d, rfl⟩
-- ⊢ a ∣ a * d + c
rcases h2 with ⟨e, rfl⟩
-- ⊢ a ∣ a * d + a * e
use (d + e)
-- ⊢ a * d + a * e = a * (d + e)
ring
-- 6ª demostración
example
(h1 : a ∣ b)
(h2 : a ∣ c)
: a ∣ (b + c) :=
dvd_add h1 h2
-- Lemas usados
-- ============
-- #check (Dvd.intro c : a * c = b → a ∣ b)
-- #check (dvd_add : a ∣ b → a ∣ c → a ∣ (b + c))
-- #check (mul_add a b c : a * (b + c) = a * b + a * c)