Theorem divcan3

Description: A cancellation law for division. (Contributed by NM, 3-Feb-2004) (Proof shortened by Mario Carneiro, 27-May-2016)

	Ref	Expression
Assertion	divcan3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		eqid	⊢ ( 𝐵 · 𝐴 ) = ( 𝐵 · 𝐴 )
2		simp2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → 𝐵 ∈ ℂ )
3		simp1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → 𝐴 ∈ ℂ )
4	2 3	mulcld	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 · 𝐴 ) ∈ ℂ )
5		3simpc	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) )
6		divmul	⊢ ( ( ( 𝐵 · 𝐴 ) ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 ↔ ( 𝐵 · 𝐴 ) = ( 𝐵 · 𝐴 ) ) )
7	4 3 5 6	syl3anc	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 ↔ ( 𝐵 · 𝐴 ) = ( 𝐵 · 𝐴 ) ) )
8	1 7	mpbiri	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 )