Metamath Proof Explorer


Theorem 2sqreunnlem2

Description: Lemma 2 for 2sqreunn . (Contributed by AV, 25-Jun-2023)

Ref Expression
Hypothesis 2sqreulem4.1 ⊢ φ ↔ ψ ∧ a 2 + b 2 = P
Assertion 2sqreunnlem2 ⊢ ∀ a ∈ ℕ ∃* b ∈ ℕ φ

Proof

Step Hyp Ref Expression
1 2sqreulem4.1 ⊢ φ ↔ ψ ∧ a 2 + b 2 = P
2 nnssnn0 ⊢ ℕ ⊆ ℕ 0
3 1 2sqreulem4 ⊢ ∀ a ∈ ℕ 0 ∃* b ∈ ℕ 0 φ
4 nfcv ⊢ Ⅎ _ b ℕ
5 nfcv ⊢ Ⅎ _ b ℕ 0
6 4 5 ssrmof ⊢ ℕ ⊆ ℕ 0 → ∃* b ∈ ℕ 0 φ → ∃* b ∈ ℕ φ
7 6 ralimdv ⊢ ℕ ⊆ ℕ 0 → ∀ a ∈ ℕ 0 ∃* b ∈ ℕ 0 φ → ∀ a ∈ ℕ 0 ∃* b ∈ ℕ φ
8 2 3 7 mp2 ⊢ ∀ a ∈ ℕ 0 ∃* b ∈ ℕ φ
9 ssralv ⊢ ℕ ⊆ ℕ 0 → ∀ a ∈ ℕ 0 ∃* b ∈ ℕ φ → ∀ a ∈ ℕ ∃* b ∈ ℕ φ
10 2 8 9 mp2 ⊢ ∀ a ∈ ℕ ∃* b ∈ ℕ φ