Metamath Proof Explorer


Theorem subrecd

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

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

Proof

Step Hyp Ref Expression
1 subrecd.1 ( 𝜑𝐴 ∈ ℂ )
2 subrecd.2 ( 𝜑𝐵 ∈ ℂ )
3 subrecd.3 ( 𝜑𝐴 ≠ 0 )
4 subrecd.4 ( 𝜑𝐵 ≠ 0 )
5 subrec ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( ( 1 / 𝐴 ) − ( 1 / 𝐵 ) ) = ( ( 𝐵𝐴 ) / ( 𝐴 · 𝐵 ) ) )
6 1 3 2 4 5 syl22anc ( 𝜑 → ( ( 1 / 𝐴 ) − ( 1 / 𝐵 ) ) = ( ( 𝐵𝐴 ) / ( 𝐴 · 𝐵 ) ) )