Metamath Proof Explorer


Theorem 0dioph

Description: The empty set is Diophantine. (Contributed by Stefan O'Rear, 10-Oct-2014)

Ref Expression
Assertion 0dioph ⊢ A ∈ ℕ 0 → ∅ ∈ Dioph ⁡ A

Proof

Step Hyp Ref Expression
1 ax-1ne0 ⊢ 1 ≠ 0
2 1 neii ⊢ ¬ 1 = 0
3 2 rgenw ⊢ ∀ a ∈ ℕ 0 1 … A ¬ 1 = 0
4 rabeq0 ⊢ a ∈ ℕ 0 1 … A | 1 = 0 = ∅ ↔ ∀ a ∈ ℕ 0 1 … A ¬ 1 = 0
5 3 4 mpbir ⊢ a ∈ ℕ 0 1 … A | 1 = 0 = ∅
6 ovex ⊢ 1 … A ∈ V
7 1z ⊢ 1 ∈ ℤ
8 mzpconstmpt ⊢ 1 … A ∈ V ∧ 1 ∈ ℤ → a ∈ ℤ 1 … A ⟼ 1 ∈ mzPoly ⁡ 1 … A
9 6 7 8 mp2an ⊢ a ∈ ℤ 1 … A ⟼ 1 ∈ mzPoly ⁡ 1 … A
10 eq0rabdioph ⊢ A ∈ ℕ 0 ∧ a ∈ ℤ 1 … A ⟼ 1 ∈ mzPoly ⁡ 1 … A → a ∈ ℕ 0 1 … A | 1 = 0 ∈ Dioph ⁡ A
11 9 10 mpan2 ⊢ A ∈ ℕ 0 → a ∈ ℕ 0 1 … A | 1 = 0 ∈ Dioph ⁡ A
12 5 11 eqeltrrid ⊢ A ∈ ℕ 0 → ∅ ∈ Dioph ⁡ A