Metamath Proof Explorer


Theorem redivmuld

Description: Relationship between division and multiplication. (Contributed by SN, 25-Nov-2025)

Ref Expression
Hypotheses redivmuld.a ( 𝜑𝐴 ∈ ℝ )
redivmuld.b ( 𝜑𝐵 ∈ ℝ )
redivmuld.c ( 𝜑𝐶 ∈ ℝ )
redivmuld.z ( 𝜑𝐶 ≠ 0 )
Assertion redivmuld ( 𝜑 → ( ( 𝐴 / 𝐶 ) = 𝐵 ↔ ( 𝐶 · 𝐵 ) = 𝐴 ) )

Proof

Step Hyp Ref Expression
1 redivmuld.a ( 𝜑𝐴 ∈ ℝ )
2 redivmuld.b ( 𝜑𝐵 ∈ ℝ )
3 redivmuld.c ( 𝜑𝐶 ∈ ℝ )
4 redivmuld.z ( 𝜑𝐶 ≠ 0 )
5 1 3 4 redivvald ( 𝜑 → ( 𝐴 / 𝐶 ) = ( 𝑥 ∈ ℝ ( 𝐶 · 𝑥 ) = 𝐴 ) )
6 5 eqeq1d ( 𝜑 → ( ( 𝐴 / 𝐶 ) = 𝐵 ↔ ( 𝑥 ∈ ℝ ( 𝐶 · 𝑥 ) = 𝐴 ) = 𝐵 ) )
7 1 3 4 rediveud ( 𝜑 → ∃! 𝑥 ∈ ℝ ( 𝐶 · 𝑥 ) = 𝐴 )
8 oveq2 ( 𝑥 = 𝐵 → ( 𝐶 · 𝑥 ) = ( 𝐶 · 𝐵 ) )
9 8 eqeq1d ( 𝑥 = 𝐵 → ( ( 𝐶 · 𝑥 ) = 𝐴 ↔ ( 𝐶 · 𝐵 ) = 𝐴 ) )
10 9 riota2 ( ( 𝐵 ∈ ℝ ∧ ∃! 𝑥 ∈ ℝ ( 𝐶 · 𝑥 ) = 𝐴 ) → ( ( 𝐶 · 𝐵 ) = 𝐴 ↔ ( 𝑥 ∈ ℝ ( 𝐶 · 𝑥 ) = 𝐴 ) = 𝐵 ) )
11 2 7 10 syl2anc ( 𝜑 → ( ( 𝐶 · 𝐵 ) = 𝐴 ↔ ( 𝑥 ∈ ℝ ( 𝐶 · 𝑥 ) = 𝐴 ) = 𝐵 ) )
12 6 11 bitr4d ( 𝜑 → ( ( 𝐴 / 𝐶 ) = 𝐵 ↔ ( 𝐶 · 𝐵 ) = 𝐴 ) )