Metamath Proof Explorer


Theorem redivcan3d

Description: A cancellation law for division. (Contributed by SN, 25-Nov-2025)

Ref Expression
Hypotheses redivcan2d.a ⊢ φ → A ∈ ℝ
redivcan2d.b ⊢ φ → B ∈ ℝ
redivcan2d.z ⊢ φ → B ≠ 0
Assertion redivcan3d ⊢ φ → B ⁢ A / ℝ B = A

Proof

Step Hyp Ref Expression
1 redivcan2d.a ⊢ φ → A ∈ ℝ
2 redivcan2d.b ⊢ φ → B ∈ ℝ
3 redivcan2d.z ⊢ φ → B ≠ 0
4 eqidd ⊢ φ → B ⁢ A = B ⁢ A
5 2 1 remulcld ⊢ φ → B ⁢ A ∈ ℝ
6 5 1 2 3 redivmuld ⊢ φ → B ⁢ A / ℝ B = A ↔ B ⁢ A = B ⁢ A
7 4 6 mpbird ⊢ φ → B ⁢ A / ℝ B = A