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Metamath Proof Explorer

Theorem nnncan

Description: Cancellation law for subtraction. (Contributed by NM, 4-Sep-2005)

Ref Expression
Assertion nnncan ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A - B − C - C = A − B

Proof

Step Hyp Ref Expression
1 subcl ⊢ B ∈ ℂ ∧ C ∈ ℂ → B − C ∈ ℂ
2 1 3adant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → B − C ∈ ℂ
3 subsub4 ⊢ A ∈ ℂ ∧ B − C ∈ ℂ ∧ C ∈ ℂ → A - B − C - C = A − B - C + C
4 2 3 syld3an2 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A - B − C - C = A − B - C + C
5 npcan ⊢ B ∈ ℂ ∧ C ∈ ℂ → B - C + C = B
6 5 oveq2d ⊢ B ∈ ℂ ∧ C ∈ ℂ → A − B - C + C = A − B
7 6 3adant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A − B - C + C = A − B
8 4 7 eqtrd ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A - B − C - C = A − B