Theorem divcan4

Description: A cancellation law for division. (Contributed by NM, 8-Feb-2005)

	Ref	Expression
Assertion	divcan4	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		mulcom	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 · 𝐵 ) = ( 𝐵 · 𝐴 ) )
2	1	3adant3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 · 𝐵 ) = ( 𝐵 · 𝐴 ) )
3	2	oveq1d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 · 𝐵 ) / 𝐵 ) = ( ( 𝐵 · 𝐴 ) / 𝐵 ) )
4		divcan3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐵 · 𝐴 ) / 𝐵 ) = 𝐴 )
5	3 4	eqtrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( ( 𝐴 · 𝐵 ) / 𝐵 ) = 𝐴 )