Metamath Proof Explorer


Theorem qdenval

Description: Value of the canonical denominator function. (Contributed by Stefan O'Rear, 13-Sep-2014)

Ref Expression
Assertion qdenval ⊢ A ∈ ℚ → denom ⁡ A = 2 nd ⁡ ι x ∈ ℤ × ℕ | 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ A = 1 st ⁡ x 2 nd ⁡ x

Proof

Step Hyp Ref Expression
1 eqeq1 ⊢ a = A → a = 1 st ⁡ x 2 nd ⁡ x ↔ A = 1 st ⁡ x 2 nd ⁡ x
2 1 anbi2d ⊢ a = A → 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ a = 1 st ⁡ x 2 nd ⁡ x ↔ 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ A = 1 st ⁡ x 2 nd ⁡ x
3 2 riotabidv ⊢ a = A → ι x ∈ ℤ × ℕ | 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ a = 1 st ⁡ x 2 nd ⁡ x = ι x ∈ ℤ × ℕ | 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ A = 1 st ⁡ x 2 nd ⁡ x
4 3 fveq2d ⊢ a = A → 2 nd ⁡ ι x ∈ ℤ × ℕ | 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ a = 1 st ⁡ x 2 nd ⁡ x = 2 nd ⁡ ι x ∈ ℤ × ℕ | 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ A = 1 st ⁡ x 2 nd ⁡ x
5 df-denom ⊢ denom = a ∈ ℚ ⟼ 2 nd ⁡ ι x ∈ ℤ × ℕ | 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ a = 1 st ⁡ x 2 nd ⁡ x
6 fvex ⊢ 2 nd ⁡ ι x ∈ ℤ × ℕ | 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ A = 1 st ⁡ x 2 nd ⁡ x ∈ V
7 4 5 6 fvmpt ⊢ A ∈ ℚ → denom ⁡ A = 2 nd ⁡ ι x ∈ ℤ × ℕ | 1 st ⁡ x gcd 2 nd ⁡ x = 1 ∧ A = 1 st ⁡ x 2 nd ⁡ x