Metamath Proof Explorer


Theorem subcanad

Description: Cancellation law for subtraction. Deduction form of subcan . Generalization of subcand . (Contributed by David Moews, 28-Feb-2017)

Ref Expression
Hypotheses negidd.1 ⊢ φ → A ∈ ℂ
pncand.2 ⊢ φ → B ∈ ℂ
subaddd.3 ⊢ φ → C ∈ ℂ
Assertion subcanad ⊢ φ → A − B = A − C ↔ B = C

Proof

Step Hyp Ref Expression
1 negidd.1 ⊢ φ → A ∈ ℂ
2 pncand.2 ⊢ φ → B ∈ ℂ
3 subaddd.3 ⊢ φ → C ∈ ℂ
4 subcan ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A − B = A − C ↔ B = C
5 1 2 3 4 syl3anc ⊢ φ → A − B = A − C ↔ B = C