Metamath Proof Explorer


Theorem 4sqlem3

Description: Lemma for 4sq . Sufficient condition to be in S . (Contributed by Mario Carneiro, 14-Jul-2014)

Ref Expression
Hypothesis 4sq.1 ⊢ S = n | ∃ x ∈ ℤ ∃ y ∈ ℤ ∃ z ∈ ℤ ∃ w ∈ ℤ n = x 2 + y 2 + z 2 + w 2
Assertion 4sqlem3 ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ C ∈ ℤ ∧ D ∈ ℤ → A 2 + B 2 + C 2 + D 2 ∈ S

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 eqid ⊢ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + C 2 + D 2
3 oveq1 ⊢ c = C → c 2 = C 2
4 3 oveq1d ⊢ c = C → c 2 + d 2 = C 2 + d 2
5 4 oveq2d ⊢ c = C → A 2 + B 2 + c 2 + d 2 = A 2 + B 2 + C 2 + d 2
6 5 eqeq2d ⊢ c = C → A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + c 2 + d 2 ↔ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + C 2 + d 2
7 oveq1 ⊢ d = D → d 2 = D 2
8 7 oveq2d ⊢ d = D → C 2 + d 2 = C 2 + D 2
9 8 oveq2d ⊢ d = D → A 2 + B 2 + C 2 + d 2 = A 2 + B 2 + C 2 + D 2
10 9 eqeq2d ⊢ d = D → A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + C 2 + d 2 ↔ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + C 2 + D 2
11 6 10 rspc2ev ⊢ C ∈ ℤ ∧ D ∈ ℤ ∧ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + C 2 + D 2 → ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + c 2 + d 2
12 2 11 mp3an3 ⊢ C ∈ ℤ ∧ D ∈ ℤ → ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + c 2 + d 2
13 oveq1 ⊢ a = A → a 2 = A 2
14 13 oveq1d ⊢ a = A → a 2 + b 2 = A 2 + b 2
15 14 oveq1d ⊢ a = A → a 2 + b 2 + c 2 + d 2 = A 2 + b 2 + c 2 + d 2
16 15 eqeq2d ⊢ a = A → A 2 + B 2 + C 2 + D 2 = a 2 + b 2 + c 2 + d 2 ↔ A 2 + B 2 + C 2 + D 2 = A 2 + b 2 + c 2 + d 2
17 16 2rexbidv ⊢ a = A → ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = a 2 + b 2 + c 2 + d 2 ↔ ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = A 2 + b 2 + c 2 + d 2
18 oveq1 ⊢ b = B → b 2 = B 2
19 18 oveq2d ⊢ b = B → A 2 + b 2 = A 2 + B 2
20 19 oveq1d ⊢ b = B → A 2 + b 2 + c 2 + d 2 = A 2 + B 2 + c 2 + d 2
21 20 eqeq2d ⊢ b = B → A 2 + B 2 + C 2 + D 2 = A 2 + b 2 + c 2 + d 2 ↔ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + c 2 + d 2
22 21 2rexbidv ⊢ b = B → ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = A 2 + b 2 + c 2 + d 2 ↔ ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + c 2 + d 2
23 17 22 rspc2ev ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + c 2 + d 2 → ∃ a ∈ ℤ ∃ b ∈ ℤ ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = a 2 + b 2 + c 2 + d 2
24 23 3expa ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + c 2 + d 2 → ∃ a ∈ ℤ ∃ b ∈ ℤ ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = a 2 + b 2 + c 2 + d 2
25 1 4sqlem2 ⊢ A 2 + B 2 + C 2 + D 2 ∈ S ↔ ∃ a ∈ ℤ ∃ b ∈ ℤ ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = a 2 + b 2 + c 2 + d 2
26 24 25 sylibr ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ ∃ c ∈ ℤ ∃ d ∈ ℤ A 2 + B 2 + C 2 + D 2 = A 2 + B 2 + c 2 + d 2 → A 2 + B 2 + C 2 + D 2 ∈ S
27 12 26 sylan2 ⊢ A ∈ ℤ ∧ B ∈ ℤ ∧ C ∈ ℤ ∧ D ∈ ℤ → A 2 + B 2 + C 2 + D 2 ∈ S