Metamath Proof Explorer
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 |
|
subrec |
⊢ ( ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) ∧ ( 𝐵 ∈ ℂ ∧ 𝐵 ≠ 0 ) ) → ( ( 1 / 𝐴 ) − ( 1 / 𝐵 ) ) = ( ( 𝐵 − 𝐴 ) / ( 𝐴 · 𝐵 ) ) ) |
6 |
1 3 2 4 5
|
syl22anc |
⊢ ( 𝜑 → ( ( 1 / 𝐴 ) − ( 1 / 𝐵 ) ) = ( ( 𝐵 − 𝐴 ) / ( 𝐴 · 𝐵 ) ) ) |