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Metamath Proof Explorer

Theorem nppcan2

Description: Cancellation law for subtraction. (Contributed by NM, 29-Sep-2005)

Ref Expression
Assertion nppcan2 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A - B + C + C = A − B

Proof

Step Hyp Ref Expression
1 addcl ⊢ B ∈ ℂ ∧ C ∈ ℂ → B + C ∈ ℂ
2 1 3adant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → B + C ∈ ℂ
3 subsub ⊢ 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 pncan ⊢ B ∈ ℂ ∧ C ∈ ℂ → B + C - C = B
6 5 3adant1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → B + C - C = B
7 6 oveq2d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A − B + C - C = A − B
8 4 7 eqtr3d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A - B + C + C = A − B