Metamath Proof Explorer


Theorem redivcan3d

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

Ref Expression
Hypotheses redivcan2d.a ( 𝜑𝐴 ∈ ℝ )
redivcan2d.b ( 𝜑𝐵 ∈ ℝ )
redivcan2d.z ( 𝜑𝐵 ≠ 0 )
Assertion redivcan3d ( 𝜑 → ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 redivcan2d.a ( 𝜑𝐴 ∈ ℝ )
2 redivcan2d.b ( 𝜑𝐵 ∈ ℝ )
3 redivcan2d.z ( 𝜑𝐵 ≠ 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 ) / 𝐵 ) = 𝐴 ↔ ( 𝐵 · 𝐴 ) = ( 𝐵 · 𝐴 ) ) )
7 4 6 mpbird ( 𝜑 → ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 )