Metamath Proof Explorer

Theorem subdivcomb1

Description: Bring a term in a subtraction into the numerator. (Contributed by Scott Fenton, 3-Jul-2013)

Ref Expression
Assertion subdivcomb1 ⊢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→C⁢A−BC=A−BC

Proof

Step Hyp Ref Expression
1 simp3l ⊢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→C∈ℂ
2 simp1 ⊢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→A∈ℂ
3 1 2 mulcld ⊢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→C⁢A∈ℂ
4 divsubdir ⊢C⁢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→C⁢A−BC=C⁢AC−BC
5 3 4 syld3an1 ⊢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→C⁢A−BC=C⁢AC−BC
6 divcan3 ⊢A∈ℂ∧C∈ℂ∧C≠0→C⁢AC=A
7 6 3expb ⊢A∈ℂ∧C∈ℂ∧C≠0→C⁢AC=A
8 7 3adant2 ⊢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→C⁢AC=A
9 8 oveq1d ⊢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→C⁢AC−BC=A−BC
10 5 9 eqtrd ⊢A∈ℂ∧B∈ℂ∧C∈ℂ∧C≠0→C⁢A−BC=A−BC