Metamath Proof Explorer


Theorem pcmpt2

Description: Dividing two prime count maps yields a number with all dividing primes confined to an interval. (Contributed by Mario Carneiro, 14-Mar-2014)

Ref Expression
Hypotheses pcmpt.1 ⊢ F = n ∈ ℕ ⟼ if n ∈ ℙ n A 1
pcmpt.2 ⊢ φ → ∀ n ∈ ℙ A ∈ ℕ 0
pcmpt.3 ⊢ φ → N ∈ ℕ
pcmpt.4 ⊢ φ → P ∈ ℙ
pcmpt.5 ⊢ n = P → A = B
pcmpt2.6 ⊢ φ → M ∈ ℤ ≥ N
Assertion pcmpt2 ⊢ φ → P pCnt seq 1 × F ⁡ M seq 1 × F ⁡ N = if P ≤ M ∧ ¬ P ≤ N B 0

Proof

Step Hyp Ref Expression
1 pcmpt.1 ⊢ F = n ∈ ℕ ⟼ if n ∈ ℙ n A 1
2 pcmpt.2 ⊢ φ → ∀ n ∈ ℙ A ∈ ℕ 0
3 pcmpt.3 ⊢ φ → N ∈ ℕ
4 pcmpt.4 ⊢ φ → P ∈ ℙ
5 pcmpt.5 ⊢ n = P → A = B
6 pcmpt2.6 ⊢ φ → M ∈ ℤ ≥ N
7 1 2 pcmptcl ⊢ φ → F : ℕ ⟶ ℕ ∧ seq 1 × F : ℕ ⟶ ℕ
8 7 simprd ⊢ φ → seq 1 × F : ℕ ⟶ ℕ
9 eluznn ⊢ N ∈ ℕ ∧ M ∈ ℤ ≥ N → M ∈ ℕ
10 3 6 9 syl2anc ⊢ φ → M ∈ ℕ
11 8 10 ffvelcdmd ⊢ φ → seq 1 × F ⁡ M ∈ ℕ
12 11 nnzd ⊢ φ → seq 1 × F ⁡ M ∈ ℤ
13 11 nnne0d ⊢ φ → seq 1 × F ⁡ M ≠ 0
14 8 3 ffvelcdmd ⊢ φ → seq 1 × F ⁡ N ∈ ℕ
15 pcdiv ⊢ P ∈ ℙ ∧ seq 1 × F ⁡ M ∈ ℤ ∧ seq 1 × F ⁡ M ≠ 0 ∧ seq 1 × F ⁡ N ∈ ℕ → P pCnt seq 1 × F ⁡ M seq 1 × F ⁡ N = P pCnt seq 1 × F ⁡ M − P pCnt seq 1 × F ⁡ N
16 4 12 13 14 15 syl121anc ⊢ φ → P pCnt seq 1 × F ⁡ M seq 1 × F ⁡ N = P pCnt seq 1 × F ⁡ M − P pCnt seq 1 × F ⁡ N
17 1 2 10 4 5 pcmpt ⊢ φ → P pCnt seq 1 × F ⁡ M = if P ≤ M B 0
18 1 2 3 4 5 pcmpt ⊢ φ → P pCnt seq 1 × F ⁡ N = if P ≤ N B 0
19 17 18 oveq12d ⊢ φ → P pCnt seq 1 × F ⁡ M − P pCnt seq 1 × F ⁡ N = if P ≤ M B 0 − if P ≤ N B 0
20 5 eleq1d ⊢ n = P → A ∈ ℕ 0 ↔ B ∈ ℕ 0
21 20 2 4 rspcdva ⊢ φ → B ∈ ℕ 0
22 21 nn0cnd ⊢ φ → B ∈ ℂ
23 22 subidd ⊢ φ → B − B = 0
24 23 adantr ⊢ φ ∧ P ≤ N → B − B = 0
25 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
26 4 25 syl ⊢ φ → P ∈ ℕ
27 26 nnred ⊢ φ → P ∈ ℝ
28 27 adantr ⊢ φ ∧ P ≤ N → P ∈ ℝ
29 3 nnred ⊢ φ → N ∈ ℝ
30 29 adantr ⊢ φ ∧ P ≤ N → N ∈ ℝ
31 10 nnred ⊢ φ → M ∈ ℝ
32 31 adantr ⊢ φ ∧ P ≤ N → M ∈ ℝ
33 simpr ⊢ φ ∧ P ≤ N → P ≤ N
34 eluzle ⊢ M ∈ ℤ ≥ N → N ≤ M
35 6 34 syl ⊢ φ → N ≤ M
36 35 adantr ⊢ φ ∧ P ≤ N → N ≤ M
37 28 30 32 33 36 letrd ⊢ φ ∧ P ≤ N → P ≤ M
38 37 iftrued ⊢ φ ∧ P ≤ N → if P ≤ M B 0 = B
39 iftrue ⊢ P ≤ N → if P ≤ N B 0 = B
40 39 adantl ⊢ φ ∧ P ≤ N → if P ≤ N B 0 = B
41 38 40 oveq12d ⊢ φ ∧ P ≤ N → if P ≤ M B 0 − if P ≤ N B 0 = B − B
42 simpr ⊢ P ≤ M ∧ ¬ P ≤ N → ¬ P ≤ N
43 42 33 nsyl3 ⊢ φ ∧ P ≤ N → ¬ P ≤ M ∧ ¬ P ≤ N
44 43 iffalsed ⊢ φ ∧ P ≤ N → if P ≤ M ∧ ¬ P ≤ N B 0 = 0
45 24 41 44 3eqtr4d ⊢ φ ∧ P ≤ N → if P ≤ M B 0 − if P ≤ N B 0 = if P ≤ M ∧ ¬ P ≤ N B 0
46 iffalse ⊢ ¬ P ≤ N → if P ≤ N B 0 = 0
47 46 oveq2d ⊢ ¬ P ≤ N → if P ≤ M B 0 − if P ≤ N B 0 = if P ≤ M B 0 − 0
48 0cn ⊢ 0 ∈ ℂ
49 ifcl ⊢ B ∈ ℂ ∧ 0 ∈ ℂ → if P ≤ M B 0 ∈ ℂ
50 22 48 49 sylancl ⊢ φ → if P ≤ M B 0 ∈ ℂ
51 50 subid1d ⊢ φ → if P ≤ M B 0 − 0 = if P ≤ M B 0
52 47 51 sylan9eqr ⊢ φ ∧ ¬ P ≤ N → if P ≤ M B 0 − if P ≤ N B 0 = if P ≤ M B 0
53 simpr ⊢ φ ∧ ¬ P ≤ N → ¬ P ≤ N
54 53 biantrud ⊢ φ ∧ ¬ P ≤ N → P ≤ M ↔ P ≤ M ∧ ¬ P ≤ N
55 54 ifbid ⊢ φ ∧ ¬ P ≤ N → if P ≤ M B 0 = if P ≤ M ∧ ¬ P ≤ N B 0
56 52 55 eqtrd ⊢ φ ∧ ¬ P ≤ N → if P ≤ M B 0 − if P ≤ N B 0 = if P ≤ M ∧ ¬ P ≤ N B 0
57 45 56 pm2.61dan ⊢ φ → if P ≤ M B 0 − if P ≤ N B 0 = if P ≤ M ∧ ¬ P ≤ N B 0
58 16 19 57 3eqtrd ⊢ φ → P pCnt seq 1 × F ⁡ M seq 1 × F ⁡ N = if P ≤ M ∧ ¬ P ≤ N B 0