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Metamath Proof Explorer

Theorem div11

Description: One-to-one relationship for division. (Contributed by NM, 20-Apr-2006) (Proof shortened by Mario Carneiro, 27-May-2016) (Proof shortened by SN, 9-Jul-2025)

Ref Expression
Assertion div11 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) ↔ 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 divcl ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( 𝐵 / 𝐶 ) ∈ ℂ )
2 1 3expb ⊢ ( ( 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐵 / 𝐶 ) ∈ ℂ )
3 2 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐵 / 𝐶 ) ∈ ℂ )
4 divmul3 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 / 𝐶 ) ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) ↔ 𝐴 = ( ( 𝐵 / 𝐶 ) · 𝐶 ) ) )
5 3 4 syld3an2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) ↔ 𝐴 = ( ( 𝐵 / 𝐶 ) · 𝐶 ) ) )
6 divcan1 ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) → ( ( 𝐵 / 𝐶 ) · 𝐶 ) = 𝐵 )
7 6 3expb ⊢ ( ( 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐵 / 𝐶 ) · 𝐶 ) = 𝐵 )
8 7 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐵 / 𝐶 ) · 𝐶 ) = 𝐵 )
9 8 eqeq2d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( 𝐴 = ( ( 𝐵 / 𝐶 ) · 𝐶 ) ↔ 𝐴 = 𝐵 ) )
10 5 9 bitrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐶 ∈ ℂ ∧ 𝐶 ≠ 0 ) ) → ( ( 𝐴 / 𝐶 ) = ( 𝐵 / 𝐶 ) ↔ 𝐴 = 𝐵 ) )