Metamath Proof Explorer


Theorem pcexp

Description: Prime power of an exponential. (Contributed by Mario Carneiro, 10-Aug-2015)

Ref Expression
Assertion pcexp ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ N ∈ ℤ → P pCnt A N = N ⁢ P pCnt A

Proof

Step Hyp Ref Expression
1 oveq2 ⊢ x = 0 → A x = A 0
2 1 oveq2d ⊢ x = 0 → P pCnt A x = P pCnt A 0
3 oveq1 ⊢ x = 0 → x ⁢ P pCnt A = 0 ⋅ P pCnt A
4 2 3 eqeq12d ⊢ x = 0 → P pCnt A x = x ⁢ P pCnt A ↔ P pCnt A 0 = 0 ⋅ P pCnt A
5 oveq2 ⊢ x = y → A x = A y
6 5 oveq2d ⊢ x = y → P pCnt A x = P pCnt A y
7 oveq1 ⊢ x = y → x ⁢ P pCnt A = y ⁢ P pCnt A
8 6 7 eqeq12d ⊢ x = y → P pCnt A x = x ⁢ P pCnt A ↔ P pCnt A y = y ⁢ P pCnt A
9 oveq2 ⊢ x = y + 1 → A x = A y + 1
10 9 oveq2d ⊢ x = y + 1 → P pCnt A x = P pCnt A y + 1
11 oveq1 ⊢ x = y + 1 → x ⁢ P pCnt A = y + 1 ⁢ P pCnt A
12 10 11 eqeq12d ⊢ x = y + 1 → P pCnt A x = x ⁢ P pCnt A ↔ P pCnt A y + 1 = y + 1 ⁢ P pCnt A
13 oveq2 ⊢ x = − y → A x = A − y
14 13 oveq2d ⊢ x = − y → P pCnt A x = P pCnt A − y
15 oveq1 ⊢ x = − y → x ⁢ P pCnt A = − y ⁢ P pCnt A
16 14 15 eqeq12d ⊢ x = − y → P pCnt A x = x ⁢ P pCnt A ↔ P pCnt A − y = − y ⁢ P pCnt A
17 oveq2 ⊢ x = N → A x = A N
18 17 oveq2d ⊢ x = N → P pCnt A x = P pCnt A N
19 oveq1 ⊢ x = N → x ⁢ P pCnt A = N ⁢ P pCnt A
20 18 19 eqeq12d ⊢ x = N → P pCnt A x = x ⁢ P pCnt A ↔ P pCnt A N = N ⁢ P pCnt A
21 pc1 ⊢ P ∈ ℙ → P pCnt 1 = 0
22 21 adantr ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → P pCnt 1 = 0
23 qcn ⊢ A ∈ ℚ → A ∈ ℂ
24 23 ad2antrl ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → A ∈ ℂ
25 24 exp0d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → A 0 = 1
26 25 oveq2d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → P pCnt A 0 = P pCnt 1
27 pcqcl ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → P pCnt A ∈ ℤ
28 27 zcnd ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → P pCnt A ∈ ℂ
29 28 mul02d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → 0 ⋅ P pCnt A = 0
30 22 26 29 3eqtr4d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → P pCnt A 0 = 0 ⋅ P pCnt A
31 oveq1 ⊢ P pCnt A y = y ⁢ P pCnt A → P pCnt A y + P pCnt A = y ⁢ P pCnt A + P pCnt A
32 expp1 ⊢ A ∈ ℂ ∧ y ∈ ℕ 0 → A y + 1 = A y ⁢ A
33 24 32 sylan ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → A y + 1 = A y ⁢ A
34 33 oveq2d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → P pCnt A y + 1 = P pCnt A y ⁢ A
35 simpll ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → P ∈ ℙ
36 simplrl ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → A ∈ ℚ
37 simplrr ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → A ≠ 0
38 nn0z ⊢ y ∈ ℕ 0 → y ∈ ℤ
39 38 adantl ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → y ∈ ℤ
40 qexpclz ⊢ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℤ → A y ∈ ℚ
41 36 37 39 40 syl3anc ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → A y ∈ ℚ
42 24 adantr ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → A ∈ ℂ
43 42 37 39 expne0d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → A y ≠ 0
44 pcqmul ⊢ P ∈ ℙ ∧ A y ∈ ℚ ∧ A y ≠ 0 ∧ A ∈ ℚ ∧ A ≠ 0 → P pCnt A y ⁢ A = P pCnt A y + P pCnt A
45 35 41 43 36 37 44 syl122anc ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → P pCnt A y ⁢ A = P pCnt A y + P pCnt A
46 34 45 eqtrd ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → P pCnt A y + 1 = P pCnt A y + P pCnt A
47 nn0cn ⊢ y ∈ ℕ 0 → y ∈ ℂ
48 47 adantl ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → y ∈ ℂ
49 28 adantr ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → P pCnt A ∈ ℂ
50 48 49 adddirp1d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → y + 1 ⁢ P pCnt A = y ⁢ P pCnt A + P pCnt A
51 46 50 eqeq12d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → P pCnt A y + 1 = y + 1 ⁢ P pCnt A ↔ P pCnt A y + P pCnt A = y ⁢ P pCnt A + P pCnt A
52 31 51 imbitrrid ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ 0 → P pCnt A y = y ⁢ P pCnt A → P pCnt A y + 1 = y + 1 ⁢ P pCnt A
53 52 ex ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → y ∈ ℕ 0 → P pCnt A y = y ⁢ P pCnt A → P pCnt A y + 1 = y + 1 ⁢ P pCnt A
54 negeq ⊢ P pCnt A y = y ⁢ P pCnt A → − P pCnt A y = − y ⁢ P pCnt A
55 nnnn0 ⊢ y ∈ ℕ → y ∈ ℕ 0
56 expneg ⊢ A ∈ ℂ ∧ y ∈ ℕ 0 → A − y = 1 A y
57 24 55 56 syl2an ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → A − y = 1 A y
58 57 oveq2d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → P pCnt A − y = P pCnt 1 A y
59 simpll ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → P ∈ ℙ
60 55 41 sylan2 ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → A y ∈ ℚ
61 55 43 sylan2 ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → A y ≠ 0
62 pcrec ⊢ P ∈ ℙ ∧ A y ∈ ℚ ∧ A y ≠ 0 → P pCnt 1 A y = − P pCnt A y
63 59 60 61 62 syl12anc ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → P pCnt 1 A y = − P pCnt A y
64 58 63 eqtrd ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → P pCnt A − y = − P pCnt A y
65 nncn ⊢ y ∈ ℕ → y ∈ ℂ
66 mulneg1 ⊢ y ∈ ℂ ∧ P pCnt A ∈ ℂ → − y ⁢ P pCnt A = − y ⁢ P pCnt A
67 65 28 66 syl2anr ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → − y ⁢ P pCnt A = − y ⁢ P pCnt A
68 64 67 eqeq12d ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → P pCnt A − y = − y ⁢ P pCnt A ↔ − P pCnt A y = − y ⁢ P pCnt A
69 54 68 imbitrrid ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ y ∈ ℕ → P pCnt A y = y ⁢ P pCnt A → P pCnt A − y = − y ⁢ P pCnt A
70 69 ex ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → y ∈ ℕ → P pCnt A y = y ⁢ P pCnt A → P pCnt A − y = − y ⁢ P pCnt A
71 4 8 12 16 20 30 53 70 zindd ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 → N ∈ ℤ → P pCnt A N = N ⁢ P pCnt A
72 71 3impia ⊢ P ∈ ℙ ∧ A ∈ ℚ ∧ A ≠ 0 ∧ N ∈ ℤ → P pCnt A N = N ⁢ P pCnt A