Metamath Proof Explorer


Theorem rediveq1d

Description: Equality in terms of unit ratio. (Contributed by SN, 2-Apr-2026)

Ref Expression
Hypotheses redivcan2d.a ⊢ φ → A ∈ ℝ
redivcan2d.b ⊢ φ → B ∈ ℝ
redivcan2d.z ⊢ φ → B ≠ 0
Assertion rediveq1d ⊢ φ → A / ℝ B = 1 ↔ A = B

Proof

Step Hyp Ref Expression
1 redivcan2d.a ⊢ φ → A ∈ ℝ
2 redivcan2d.b ⊢ φ → B ∈ ℝ
3 redivcan2d.z ⊢ φ → B ≠ 0
4 1red ⊢ φ → 1 ∈ ℝ
5 1 4 2 3 redivmul2d ⊢ φ → A / ℝ B = 1 ↔ A = B ⋅ 1
6 ax-1rid ⊢ B ∈ ℝ → B ⋅ 1 = B
7 2 6 syl ⊢ φ → B ⋅ 1 = B
8 7 eqeq2d ⊢ φ → A = B ⋅ 1 ↔ A = B
9 5 8 bitrd ⊢ φ → A / ℝ B = 1 ↔ A = B