Metamath Proof Explorer


Theorem 4sqlem13

Description: Lemma for 4sq . (Contributed by Mario Carneiro, 16-Jul-2014) (Revised by AV, 14-Sep-2020)

Ref Expression
Hypotheses 4sq.1 ⊢ S = n | ∃ x ∈ ℤ ∃ y ∈ ℤ ∃ z ∈ ℤ ∃ w ∈ ℤ n = x 2 + y 2 + z 2 + w 2
4sq.2 ⊢ φ → N ∈ ℕ
4sq.3 ⊢ φ → P = 2 ⋅ N + 1
4sq.4 ⊢ φ → P ∈ ℙ
4sq.5 ⊢ φ → 0 … 2 ⋅ N ⊆ S
4sq.6 ⊢ T = i ∈ ℕ | i ⁢ P ∈ S
4sq.7 ⊢ M = inf T ℝ <
Assertion 4sqlem13 ⊢ φ → T ≠ ∅ ∧ M < P

Proof

Step Hyp Ref Expression
1 4sq.1 ⊢ S = n | ∃ x ∈ ℤ ∃ y ∈ ℤ ∃ z ∈ ℤ ∃ w ∈ ℤ n = x 2 + y 2 + z 2 + w 2
2 4sq.2 ⊢ φ → N ∈ ℕ
3 4sq.3 ⊢ φ → P = 2 ⋅ N + 1
4 4sq.4 ⊢ φ → P ∈ ℙ
5 4sq.5 ⊢ φ → 0 … 2 ⋅ N ⊆ S
6 4sq.6 ⊢ T = i ∈ ℕ | i ⁢ P ∈ S
7 4sq.7 ⊢ M = inf T ℝ <
8 eqid ⊢ u | ∃ m ∈ 0 … N u = m 2 mod P = u | ∃ m ∈ 0 … N u = m 2 mod P
9 eqid ⊢ v ∈ u | ∃ m ∈ 0 … N u = m 2 mod P ⟼ P - 1 - v = v ∈ u | ∃ m ∈ 0 … N u = m 2 mod P ⟼ P - 1 - v
10 1 2 3 4 8 9 4sqlem12 ⊢ φ → ∃ k ∈ 1 … P − 1 ∃ u ∈ ℤ i u 2 + 1 = k ⁢ P
11 simplrl ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → k ∈ 1 … P − 1
12 elfznn ⊢ k ∈ 1 … P − 1 → k ∈ ℕ
13 11 12 syl ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → k ∈ ℕ
14 simpr ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → u 2 + 1 = k ⁢ P
15 abs1 ⊢ 1 = 1
16 15 oveq1i ⊢ 1 2 = 1 2
17 sq1 ⊢ 1 2 = 1
18 16 17 eqtri ⊢ 1 2 = 1
19 18 oveq2i ⊢ u 2 + 1 2 = u 2 + 1
20 simplrr ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → u ∈ ℤ i
21 1z ⊢ 1 ∈ ℤ
22 zgz ⊢ 1 ∈ ℤ → 1 ∈ ℤ i
23 21 22 ax-mp ⊢ 1 ∈ ℤ i
24 1 4sqlem4a ⊢ u ∈ ℤ i ∧ 1 ∈ ℤ i → u 2 + 1 2 ∈ S
25 20 23 24 sylancl ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → u 2 + 1 2 ∈ S
26 19 25 eqeltrrid ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → u 2 + 1 ∈ S
27 14 26 eqeltrrd ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → k ⁢ P ∈ S
28 oveq1 ⊢ i = k → i ⁢ P = k ⁢ P
29 28 eleq1d ⊢ i = k → i ⁢ P ∈ S ↔ k ⁢ P ∈ S
30 29 6 elrab2 ⊢ k ∈ T ↔ k ∈ ℕ ∧ k ⁢ P ∈ S
31 13 27 30 sylanbrc ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → k ∈ T
32 31 ne0d ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → T ≠ ∅
33 6 ssrab3 ⊢ T ⊆ ℕ
34 nnuz ⊢ ℕ = ℤ ≥ 1
35 33 34 sseqtri ⊢ T ⊆ ℤ ≥ 1
36 infssuzcl ⊢ T ⊆ ℤ ≥ 1 ∧ T ≠ ∅ → inf T ℝ < ∈ T
37 35 32 36 sylancr ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → inf T ℝ < ∈ T
38 7 37 eqeltrid ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → M ∈ T
39 33 38 sselid ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → M ∈ ℕ
40 39 nnred ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → M ∈ ℝ
41 13 nnred ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → k ∈ ℝ
42 4 ad2antrr ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → P ∈ ℙ
43 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
44 42 43 syl ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → P ∈ ℕ
45 44 nnred ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → P ∈ ℝ
46 infssuzle ⊢ T ⊆ ℤ ≥ 1 ∧ k ∈ T → inf T ℝ < ≤ k
47 35 31 46 sylancr ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → inf T ℝ < ≤ k
48 7 47 eqbrtrid ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → M ≤ k
49 prmz ⊢ P ∈ ℙ → P ∈ ℤ
50 42 49 syl ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → P ∈ ℤ
51 elfzm11 ⊢ 1 ∈ ℤ ∧ P ∈ ℤ → k ∈ 1 … P − 1 ↔ k ∈ ℤ ∧ 1 ≤ k ∧ k < P
52 21 50 51 sylancr ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → k ∈ 1 … P − 1 ↔ k ∈ ℤ ∧ 1 ≤ k ∧ k < P
53 11 52 mpbid ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → k ∈ ℤ ∧ 1 ≤ k ∧ k < P
54 53 simp3d ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → k < P
55 40 41 45 48 54 lelttrd ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → M < P
56 32 55 jca ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i ∧ u 2 + 1 = k ⁢ P → T ≠ ∅ ∧ M < P
57 56 ex ⊢ φ ∧ k ∈ 1 … P − 1 ∧ u ∈ ℤ i → u 2 + 1 = k ⁢ P → T ≠ ∅ ∧ M < P
58 57 rexlimdvva ⊢ φ → ∃ k ∈ 1 … P − 1 ∃ u ∈ ℤ i u 2 + 1 = k ⁢ P → T ≠ ∅ ∧ M < P
59 10 58 mpd ⊢ φ → T ≠ ∅ ∧ M < P