Metamath Proof Explorer


Theorem rediv11d

Description: One-to-one relationship for division. (Contributed by SN, 9-Apr-2026)

Ref Expression
Hypotheses rediv23d.a ( 𝜑𝐴 ∈ ℝ )
rediv23d.b ( 𝜑𝐵 ∈ ℝ )
rediv23d.c ( 𝜑𝐶 ∈ ℝ )
rediv23d.z ( 𝜑𝐶 ≠ 0 )
Assertion rediv11d ( 𝜑 → ( ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) ↔ 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 rediv23d.a ( 𝜑𝐴 ∈ ℝ )
2 rediv23d.b ( 𝜑𝐵 ∈ ℝ )
3 rediv23d.c ( 𝜑𝐶 ∈ ℝ )
4 rediv23d.z ( 𝜑𝐶 ≠ 0 )
5 2 3 4 sn-redivcld ( 𝜑 → ( 𝐵 / 𝐶 ) ∈ ℝ )
6 1 5 3 4 redivmul2d ( 𝜑 → ( ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) ↔ 𝐴 = ( 𝐶 · ( 𝐵 / 𝐶 ) ) ) )
7 2 3 4 redivcan2d ( 𝜑 → ( 𝐶 · ( 𝐵 / 𝐶 ) ) = 𝐵 )
8 7 eqeq2d ( 𝜑 → ( 𝐴 = ( 𝐶 · ( 𝐵 / 𝐶 ) ) ↔ 𝐴 = 𝐵 ) )
9 6 8 bitrd ( 𝜑 → ( ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) ↔ 𝐴 = 𝐵 ) )