Metamath Proof Explorer


Theorem eulerthlem1

Description: Lemma for eulerth . (Contributed by Mario Carneiro, 8-May-2015)

Ref Expression
Hypotheses eulerth.1 ⊢ φ → N ∈ ℕ ∧ A ∈ ℤ ∧ A gcd N = 1
eulerth.2 ⊢ S = y ∈ 0 ..^ N | y gcd N = 1
eulerth.3 ⊢ T = 1 … ϕ ⁡ N
eulerth.4 ⊢ φ → F : T ⟶ 1-1 onto S
eulerth.5 ⊢ G = x ∈ T ⟼ A ⁢ F ⁡ x mod N
Assertion eulerthlem1 ⊢ φ → G : T ⟶ S

Proof

Step Hyp Ref Expression
1 eulerth.1 ⊢ φ → N ∈ ℕ ∧ A ∈ ℤ ∧ A gcd N = 1
2 eulerth.2 ⊢ S = y ∈ 0 ..^ N | y gcd N = 1
3 eulerth.3 ⊢ T = 1 … ϕ ⁡ N
4 eulerth.4 ⊢ φ → F : T ⟶ 1-1 onto S
5 eulerth.5 ⊢ G = x ∈ T ⟼ A ⁢ F ⁡ x mod N
6 1 simp2d ⊢ φ → A ∈ ℤ
7 6 adantr ⊢ φ ∧ x ∈ T → A ∈ ℤ
8 f1of ⊢ F : T ⟶ 1-1 onto S → F : T ⟶ S
9 4 8 syl ⊢ φ → F : T ⟶ S
10 9 ffvelcdmda ⊢ φ ∧ x ∈ T → F ⁡ x ∈ S
11 oveq1 ⊢ y = F ⁡ x → y gcd N = F ⁡ x gcd N
12 11 eqeq1d ⊢ y = F ⁡ x → y gcd N = 1 ↔ F ⁡ x gcd N = 1
13 12 2 elrab2 ⊢ F ⁡ x ∈ S ↔ F ⁡ x ∈ 0 ..^ N ∧ F ⁡ x gcd N = 1
14 10 13 sylib ⊢ φ ∧ x ∈ T → F ⁡ x ∈ 0 ..^ N ∧ F ⁡ x gcd N = 1
15 14 simpld ⊢ φ ∧ x ∈ T → F ⁡ x ∈ 0 ..^ N
16 elfzoelz ⊢ F ⁡ x ∈ 0 ..^ N → F ⁡ x ∈ ℤ
17 15 16 syl ⊢ φ ∧ x ∈ T → F ⁡ x ∈ ℤ
18 7 17 zmulcld ⊢ φ ∧ x ∈ T → A ⁢ F ⁡ x ∈ ℤ
19 1 simp1d ⊢ φ → N ∈ ℕ
20 19 adantr ⊢ φ ∧ x ∈ T → N ∈ ℕ
21 zmodfzo ⊢ A ⁢ F ⁡ x ∈ ℤ ∧ N ∈ ℕ → A ⁢ F ⁡ x mod N ∈ 0 ..^ N
22 18 20 21 syl2anc ⊢ φ ∧ x ∈ T → A ⁢ F ⁡ x mod N ∈ 0 ..^ N
23 modgcd ⊢ A ⁢ F ⁡ x ∈ ℤ ∧ N ∈ ℕ → A ⁢ F ⁡ x mod N gcd N = A ⁢ F ⁡ x gcd N
24 18 20 23 syl2anc ⊢ φ ∧ x ∈ T → A ⁢ F ⁡ x mod N gcd N = A ⁢ F ⁡ x gcd N
25 19 nnzd ⊢ φ → N ∈ ℤ
26 25 adantr ⊢ φ ∧ x ∈ T → N ∈ ℤ
27 18 26 gcdcomd ⊢ φ ∧ x ∈ T → A ⁢ F ⁡ x gcd N = N gcd A ⁢ F ⁡ x
28 25 6 gcdcomd ⊢ φ → N gcd A = A gcd N
29 1 simp3d ⊢ φ → A gcd N = 1
30 28 29 eqtrd ⊢ φ → N gcd A = 1
31 30 adantr ⊢ φ ∧ x ∈ T → N gcd A = 1
32 26 17 gcdcomd ⊢ φ ∧ x ∈ T → N gcd F ⁡ x = F ⁡ x gcd N
33 14 simprd ⊢ φ ∧ x ∈ T → F ⁡ x gcd N = 1
34 32 33 eqtrd ⊢ φ ∧ x ∈ T → N gcd F ⁡ x = 1
35 rpmul ⊢ N ∈ ℤ ∧ A ∈ ℤ ∧ F ⁡ x ∈ ℤ → N gcd A = 1 ∧ N gcd F ⁡ x = 1 → N gcd A ⁢ F ⁡ x = 1
36 26 7 17 35 syl3anc ⊢ φ ∧ x ∈ T → N gcd A = 1 ∧ N gcd F ⁡ x = 1 → N gcd A ⁢ F ⁡ x = 1
37 31 34 36 mp2and ⊢ φ ∧ x ∈ T → N gcd A ⁢ F ⁡ x = 1
38 24 27 37 3eqtrd ⊢ φ ∧ x ∈ T → A ⁢ F ⁡ x mod N gcd N = 1
39 oveq1 ⊢ y = A ⁢ F ⁡ x mod N → y gcd N = A ⁢ F ⁡ x mod N gcd N
40 39 eqeq1d ⊢ y = A ⁢ F ⁡ x mod N → y gcd N = 1 ↔ A ⁢ F ⁡ x mod N gcd N = 1
41 40 2 elrab2 ⊢ A ⁢ F ⁡ x mod N ∈ S ↔ A ⁢ F ⁡ x mod N ∈ 0 ..^ N ∧ A ⁢ F ⁡ x mod N gcd N = 1
42 22 38 41 sylanbrc ⊢ φ ∧ x ∈ T → A ⁢ F ⁡ x mod N ∈ S
43 42 5 fmptd ⊢ φ → G : T ⟶ S