Metamath Proof Explorer


Theorem redivrec2d

Description: Relationship between division and reciprocal. (Contributed by SN, 9-Apr-2026)

Ref Expression
Hypotheses redivrec2d.a ⊢ φ → A ∈ ℝ
redivrec2d.b ⊢ φ → B ∈ ℝ
redivrec2d.z ⊢ φ → B ≠ 0
Assertion redivrec2d ⊢ φ → A / ℝ B = 1 / ℝ B ⁢ A

Proof

Step Hyp Ref Expression
1 redivrec2d.a ⊢ φ → A ∈ ℝ
2 redivrec2d.b ⊢ φ → B ∈ ℝ
3 redivrec2d.z ⊢ φ → B ≠ 0
4 2 3 rerecidd ⊢ φ → B ⁢ 1 / ℝ B = 1
5 4 oveq1d ⊢ φ → B ⁢ 1 / ℝ B ⁢ A = 1 ⁢ A
6 2 recnd ⊢ φ → B ∈ ℂ
7 2 3 sn-rereccld ⊢ φ → 1 / ℝ B ∈ ℝ
8 7 recnd ⊢ φ → 1 / ℝ B ∈ ℂ
9 1 recnd ⊢ φ → A ∈ ℂ
10 6 8 9 mulassd ⊢ φ → B ⁢ 1 / ℝ B ⁢ A = B ⁢ 1 / ℝ B ⁢ A
11 remullid ⊢ A ∈ ℝ → 1 ⁢ A = A
12 1 11 syl ⊢ φ → 1 ⁢ A = A
13 5 10 12 3eqtr3d ⊢ φ → B ⁢ 1 / ℝ B ⁢ A = A
14 7 1 remulcld ⊢ φ → 1 / ℝ B ⁢ A ∈ ℝ
15 1 14 2 3 redivmuld ⊢ φ → A / ℝ B = 1 / ℝ B ⁢ A ↔ B ⁢ 1 / ℝ B ⁢ A = A
16 13 15 mpbird ⊢ φ → A / ℝ B = 1 / ℝ B ⁢ A