Metamath Proof Explorer


Theorem remulscllem1

Description: Lemma for remulscl . Split a product of reciprocals of naturals. (Contributed by Scott Fenton, 16-Apr-2025)

Ref Expression
Assertion remulscllem1 ( ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) ↔ ∃ 𝑛 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) )

Proof

Step Hyp Ref Expression
1 oveq2 ⊢ ( 𝑛 = ( 𝑝 ·s 𝑞 ) → ( 1s /su 𝑛 ) = ( 1s /su ( 𝑝 ·s 𝑞 ) ) )
2 1 oveq2d ⊢ ( 𝑛 = ( 𝑝 ·s 𝑞 ) → ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( 1s /su ( 𝑝 ·s 𝑞 ) ) ) )
3 2 eqeq2d ⊢ ( 𝑛 = ( 𝑝 ·s 𝑞 ) → ( ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) ↔ ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) = ( 𝐵 𝐹 ( 1s /su ( 𝑝 ·s 𝑞 ) ) ) ) )
4 nnmulscl ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → ( 𝑝 ·s 𝑞 ) ∈ ℕs )
5 1no ⊢ 1s ∈ No
6 5 a1i ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → 1s ∈ No )
7 nnno ⊢ ( 𝑝 ∈ ℕs → 𝑝 ∈ No )
8 7 adantr ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → 𝑝 ∈ No )
9 nnno ⊢ ( 𝑞 ∈ ℕs → 𝑞 ∈ No )
10 9 adantl ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → 𝑞 ∈ No )
11 nnne0s ⊢ ( 𝑝 ∈ ℕs → 𝑝 ≠ 0s )
12 11 adantr ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → 𝑝 ≠ 0s )
13 nnne0s ⊢ ( 𝑞 ∈ ℕs → 𝑞 ≠ 0s )
14 13 adantl ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → 𝑞 ≠ 0s )
15 6 8 6 10 12 14 divmuldivsd ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) = ( ( 1s ·s 1s ) /su ( 𝑝 ·s 𝑞 ) ) )
16 mulsrid ⊢ ( 1s ∈ No → ( 1s ·s 1s ) = 1s )
17 5 16 ax-mp ⊢ ( 1s ·s 1s ) = 1s
18 17 oveq1i ⊢ ( ( 1s ·s 1s ) /su ( 𝑝 ·s 𝑞 ) ) = ( 1s /su ( 𝑝 ·s 𝑞 ) )
19 15 18 eqtrdi ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) = ( 1s /su ( 𝑝 ·s 𝑞 ) ) )
20 19 oveq2d ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) = ( 𝐵 𝐹 ( 1s /su ( 𝑝 ·s 𝑞 ) ) ) )
21 3 4 20 rspcedvdw ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → ∃ 𝑛 ∈ ℕs ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) )
22 eqeq1 ⊢ ( 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) → ( 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) ↔ ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) ) )
23 22 rexbidv ⊢ ( 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) → ( ∃ 𝑛 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) ↔ ∃ 𝑛 ∈ ℕs ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) ) )
24 21 23 syl5ibrcom ⊢ ( ( 𝑝 ∈ ℕs ∧ 𝑞 ∈ ℕs ) → ( 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) → ∃ 𝑛 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) ) )
25 24 rexlimivv ⊢ ( ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) → ∃ 𝑛 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) )
26 5 a1i ⊢ ( 𝑛 ∈ ℕs → 1s ∈ No )
27 nnno ⊢ ( 𝑛 ∈ ℕs → 𝑛 ∈ No )
28 nnne0s ⊢ ( 𝑛 ∈ ℕs → 𝑛 ≠ 0s )
29 26 27 28 divscld ⊢ ( 𝑛 ∈ ℕs → ( 1s /su 𝑛 ) ∈ No )
30 29 mulsridd ⊢ ( 𝑛 ∈ ℕs → ( ( 1s /su 𝑛 ) ·s 1s ) = ( 1s /su 𝑛 ) )
31 30 eqcomd ⊢ ( 𝑛 ∈ ℕs → ( 1s /su 𝑛 ) = ( ( 1s /su 𝑛 ) ·s 1s ) )
32 31 oveq2d ⊢ ( 𝑛 ∈ ℕs → ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s 1s ) ) )
33 1nns ⊢ 1s ∈ ℕs
34 oveq2 ⊢ ( 𝑝 = 𝑛 → ( 1s /su 𝑝 ) = ( 1s /su 𝑛 ) )
35 34 oveq1d ⊢ ( 𝑝 = 𝑛 → ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) = ( ( 1s /su 𝑛 ) ·s ( 1s /su 𝑞 ) ) )
36 35 oveq2d ⊢ ( 𝑝 = 𝑛 → ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s ( 1s /su 𝑞 ) ) ) )
37 36 eqeq2d ⊢ ( 𝑝 = 𝑛 → ( ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) ↔ ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s ( 1s /su 𝑞 ) ) ) ) )
38 oveq2 ⊢ ( 𝑞 = 1s → ( 1s /su 𝑞 ) = ( 1s /su 1s ) )
39 divs1 ⊢ ( 1s ∈ No → ( 1s /su 1s ) = 1s )
40 5 39 ax-mp ⊢ ( 1s /su 1s ) = 1s
41 38 40 eqtrdi ⊢ ( 𝑞 = 1s → ( 1s /su 𝑞 ) = 1s )
42 41 oveq2d ⊢ ( 𝑞 = 1s → ( ( 1s /su 𝑛 ) ·s ( 1s /su 𝑞 ) ) = ( ( 1s /su 𝑛 ) ·s 1s ) )
43 42 oveq2d ⊢ ( 𝑞 = 1s → ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s ( 1s /su 𝑞 ) ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s 1s ) ) )
44 43 eqeq2d ⊢ ( 𝑞 = 1s → ( ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s ( 1s /su 𝑞 ) ) ) ↔ ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s 1s ) ) ) )
45 37 44 rspc2ev ⊢ ( ( 𝑛 ∈ ℕs ∧ 1s ∈ ℕs ∧ ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s 1s ) ) ) → ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) )
46 33 45 mp3an2 ⊢ ( ( 𝑛 ∈ ℕs ∧ ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑛 ) ·s 1s ) ) ) → ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) )
47 32 46 mpdan ⊢ ( 𝑛 ∈ ℕs → ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) )
48 eqeq1 ⊢ ( 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) → ( 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) ↔ ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) ) )
49 48 2rexbidv ⊢ ( 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) → ( ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) ↔ ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) ) )
50 47 49 syl5ibrcom ⊢ ( 𝑛 ∈ ℕs → ( 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) → ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) ) )
51 50 rexlimiv ⊢ ( ∃ 𝑛 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) → ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) )
52 25 51 impbii ⊢ ( ∃ 𝑝 ∈ ℕs ∃ 𝑞 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( ( 1s /su 𝑝 ) ·s ( 1s /su 𝑞 ) ) ) ↔ ∃ 𝑛 ∈ ℕs 𝐴 = ( 𝐵 𝐹 ( 1s /su 𝑛 ) ) )