Theorem divcan1

Description: A cancellation law for division. (Contributed by NM, 5-Jun-2004) (Revised by Mario Carneiro, 27-May-2016)

	Ref	Expression
Assertion	divcan1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) · 𝐵 ) = 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		divcl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) ∈ ℂ )
2		simp2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → 𝐵 ∈ ℂ )
3	1 2	mulcomd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) · 𝐵 ) = ( 𝐵 · ( 𝐴 / 𝐵 ) ) )
4		divcan2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 · ( 𝐴 / 𝐵 ) ) = 𝐴 )
5	3 4	eqtrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 / 𝐵 ) · 𝐵 ) = 𝐴 )