Metamath Proof Explorer


Theorem ablfac1lem

Description: Lemma for ablfac1b . Satisfy the assumptions of ablfacrp. (Contributed by Mario Carneiro, 26-Apr-2016)

Ref Expression
Hypotheses ablfac1.b ⊢ B = Base G
ablfac1.o ⊢ O = od ⁡ G
ablfac1.s ⊢ S = p ∈ A ⟼ x ∈ B | O ⁡ x ∥ p p pCnt B
ablfac1.g ⊢ φ → G ∈ Abel
ablfac1.f ⊢ φ → B ∈ Fin
ablfac1.1 ⊢ φ → A ⊆ ℙ
ablfac1.m ⊢ M = P P pCnt B
ablfac1.n ⊢ N = B M
Assertion ablfac1lem ⊢ φ ∧ P ∈ A → M ∈ ℕ ∧ N ∈ ℕ ∧ M gcd N = 1 ∧ B = M ⋅ N

Proof

Step Hyp Ref Expression
1 ablfac1.b ⊢ B = Base G
2 ablfac1.o ⊢ O = od ⁡ G
3 ablfac1.s ⊢ S = p ∈ A ⟼ x ∈ B | O ⁡ x ∥ p p pCnt B
4 ablfac1.g ⊢ φ → G ∈ Abel
5 ablfac1.f ⊢ φ → B ∈ Fin
6 ablfac1.1 ⊢ φ → A ⊆ ℙ
7 ablfac1.m ⊢ M = P P pCnt B
8 ablfac1.n ⊢ N = B M
9 6 sselda ⊢ φ ∧ P ∈ A → P ∈ ℙ
10 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
11 9 10 syl ⊢ φ ∧ P ∈ A → P ∈ ℕ
12 ablgrp ⊢ G ∈ Abel → G ∈ Grp
13 1 grpbn0 ⊢ G ∈ Grp → B ≠ ∅
14 4 12 13 3syl ⊢ φ → B ≠ ∅
15 hashnncl ⊢ B ∈ Fin → B ∈ ℕ ↔ B ≠ ∅
16 5 15 syl ⊢ φ → B ∈ ℕ ↔ B ≠ ∅
17 14 16 mpbird ⊢ φ → B ∈ ℕ
18 17 adantr ⊢ φ ∧ P ∈ A → B ∈ ℕ
19 9 18 pccld ⊢ φ ∧ P ∈ A → P pCnt B ∈ ℕ 0
20 11 19 nnexpcld ⊢ φ ∧ P ∈ A → P P pCnt B ∈ ℕ
21 7 20 eqeltrid ⊢ φ ∧ P ∈ A → M ∈ ℕ
22 pcdvds ⊢ P ∈ ℙ ∧ B ∈ ℕ → P P pCnt B ∥ B
23 9 18 22 syl2anc ⊢ φ ∧ P ∈ A → P P pCnt B ∥ B
24 7 23 eqbrtrid ⊢ φ ∧ P ∈ A → M ∥ B
25 nndivdvds ⊢ B ∈ ℕ ∧ M ∈ ℕ → M ∥ B ↔ B M ∈ ℕ
26 18 21 25 syl2anc ⊢ φ ∧ P ∈ A → M ∥ B ↔ B M ∈ ℕ
27 24 26 mpbid ⊢ φ ∧ P ∈ A → B M ∈ ℕ
28 8 27 eqeltrid ⊢ φ ∧ P ∈ A → N ∈ ℕ
29 21 28 jca ⊢ φ ∧ P ∈ A → M ∈ ℕ ∧ N ∈ ℕ
30 7 oveq1i ⊢ M gcd N = P P pCnt B gcd N
31 pcndvds2 ⊢ P ∈ ℙ ∧ B ∈ ℕ → ¬ P ∥ B P P pCnt B
32 9 18 31 syl2anc ⊢ φ ∧ P ∈ A → ¬ P ∥ B P P pCnt B
33 7 oveq2i ⊢ B M = B P P pCnt B
34 8 33 eqtri ⊢ N = B P P pCnt B
35 34 breq2i ⊢ P ∥ N ↔ P ∥ B P P pCnt B
36 32 35 sylnibr ⊢ φ ∧ P ∈ A → ¬ P ∥ N
37 28 nnzd ⊢ φ ∧ P ∈ A → N ∈ ℤ
38 coprm ⊢ P ∈ ℙ ∧ N ∈ ℤ → ¬ P ∥ N ↔ P gcd N = 1
39 9 37 38 syl2anc ⊢ φ ∧ P ∈ A → ¬ P ∥ N ↔ P gcd N = 1
40 36 39 mpbid ⊢ φ ∧ P ∈ A → P gcd N = 1
41 prmz ⊢ P ∈ ℙ → P ∈ ℤ
42 9 41 syl ⊢ φ ∧ P ∈ A → P ∈ ℤ
43 rpexp1i ⊢ P ∈ ℤ ∧ N ∈ ℤ ∧ P pCnt B ∈ ℕ 0 → P gcd N = 1 → P P pCnt B gcd N = 1
44 42 37 19 43 syl3anc ⊢ φ ∧ P ∈ A → P gcd N = 1 → P P pCnt B gcd N = 1
45 40 44 mpd ⊢ φ ∧ P ∈ A → P P pCnt B gcd N = 1
46 30 45 eqtrid ⊢ φ ∧ P ∈ A → M gcd N = 1
47 8 oveq2i ⊢ M ⋅ N = M ⁢ B M
48 18 nncnd ⊢ φ ∧ P ∈ A → B ∈ ℂ
49 21 nncnd ⊢ φ ∧ P ∈ A → M ∈ ℂ
50 21 nnne0d ⊢ φ ∧ P ∈ A → M ≠ 0
51 48 49 50 divcan2d ⊢ φ ∧ P ∈ A → M ⁢ B M = B
52 47 51 eqtr2id ⊢ φ ∧ P ∈ A → B = M ⋅ N
53 29 46 52 3jca ⊢ φ ∧ P ∈ A → M ∈ ℕ ∧ N ∈ ℕ ∧ M gcd N = 1 ∧ B = M ⋅ N