Theorem subrecd

Description: Subtraction of reciprocals. (Contributed by Scott Fenton, 9-Jan-2017)

	Ref	Expression
Hypotheses	subrecd.1	⊢ ( 𝜑 → 𝐴 ∈ ℂ )
	subrecd.2	⊢ ( 𝜑 → 𝐵 ∈ ℂ )
	subrecd.3	⊢ ( 𝜑 → 𝐴 ≠ 0 )
	subrecd.4	⊢ ( 𝜑 → 𝐵 ≠ 0 )
Assertion	subrecd	⊢ ( 𝜑 → ( ( 1 / 𝐴 ) − ( 1 / 𝐵 ) ) = ( ( 𝐵 − 𝐴 ) / ( 𝐴 · 𝐵 ) ) )

Proof

Step	Hyp	Ref	Expression
1		subrecd.1	⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2		subrecd.2	⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3		subrecd.3	⊢ ( 𝜑 → 𝐴 ≠ 0 )
4		subrecd.4	⊢ ( 𝜑 → 𝐵 ≠ 0 )
5		1cnd	⊢ ( 𝜑 → 1 ∈ ℂ )
6	5 1 5 2 3 4	divsubdivd	⊢ ( 𝜑 → ( ( 1 / 𝐴 ) − ( 1 / 𝐵 ) ) = ( ( ( 1 · 𝐵 ) − ( 1 · 𝐴 ) ) / ( 𝐴 · 𝐵 ) ) )
7	2	mullidd	⊢ ( 𝜑 → ( 1 · 𝐵 ) = 𝐵 )
8	1	mullidd	⊢ ( 𝜑 → ( 1 · 𝐴 ) = 𝐴 )
9	7 8	oveq12d	⊢ ( 𝜑 → ( ( 1 · 𝐵 ) − ( 1 · 𝐴 ) ) = ( 𝐵 − 𝐴 ) )
10	9	oveq1d	⊢ ( 𝜑 → ( ( ( 1 · 𝐵 ) − ( 1 · 𝐴 ) ) / ( 𝐴 · 𝐵 ) ) = ( ( 𝐵 − 𝐴 ) / ( 𝐴 · 𝐵 ) ) )
11	6 10	eqtrd	⊢ ( 𝜑 → ( ( 1 / 𝐴 ) − ( 1 / 𝐵 ) ) = ( ( 𝐵 − 𝐴 ) / ( 𝐴 · 𝐵 ) ) )