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Metamath Proof Explorer

Theorem subcan

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005) (Revised by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion subcan ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A − B = A − C ↔ B = C

Proof

Step Hyp Ref Expression
1 simp2 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → B ∈ ℂ
2 simp1 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A ∈ ℂ
3 1 2 addcomd ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → B + A = A + B
4 3 eqeq1d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → B + A = A + C ↔ A + B = A + C
5 simp3 ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → C ∈ ℂ
6 addsubeq4 ⊢ B ∈ ℂ ∧ A ∈ ℂ ∧ A ∈ ℂ ∧ C ∈ ℂ → B + A = A + C ↔ A − B = A − C
7 1 2 2 5 6 syl22anc ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → B + A = A + C ↔ A − B = A − C
8 addcan ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A + B = A + C ↔ B = C
9 4 7 8 3bitr3d ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A − B = A − C ↔ B = C