Metamath Proof Explorer


Theorem redivmuld

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

Ref Expression
Hypotheses redivmuld.a ⊢ φ → A ∈ ℝ
redivmuld.b ⊢ φ → B ∈ ℝ
redivmuld.c ⊢ φ → C ∈ ℝ
redivmuld.z ⊢ φ → C ≠ 0
Assertion redivmuld ⊢ φ → A / ℝ C = B ↔ C ⁢ B = A

Proof

Step Hyp Ref Expression
1 redivmuld.a ⊢ φ → A ∈ ℝ
2 redivmuld.b ⊢ φ → B ∈ ℝ
3 redivmuld.c ⊢ φ → C ∈ ℝ
4 redivmuld.z ⊢ φ → C ≠ 0
5 1 3 4 redivvald ⊢ φ → A / ℝ C = ι x ∈ ℝ | C ⁢ x = A
6 5 eqeq1d ⊢ φ → A / ℝ C = B ↔ ι 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 ↔ ι x ∈ ℝ | C ⁢ x = A = B
11 2 7 10 syl2anc ⊢ φ → C ⁢ B = A ↔ ι x ∈ ℝ | C ⁢ x = A = B
12 6 11 bitr4d ⊢ φ → A / ℝ C = B ↔ C ⁢ B = A