Theorem divrec2

Description: Relationship between division and reciprocal. (Contributed by NM, 7-Feb-2006)

	Ref	Expression
Assertion	divrec2	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( ( 1 / 𝐵 ) · 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		divrec	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( 𝐴 · ( 1 / 𝐵 ) ) )
2		simp1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → 𝐴 ∈ ℂ )
3		reccl	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 1 / 𝐵 ) ∈ ℂ )
4	3	3adant1	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 1 / 𝐵 ) ∈ ℂ )
5	2 4	mulcomd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 · ( 1 / 𝐵 ) ) = ( ( 1 / 𝐵 ) · 𝐴 ) )
6	1 5	eqtrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) → ( 𝐴 / 𝐵 ) = ( ( 1 / 𝐵 ) · 𝐴 ) )