Metamath Proof Explorer


Theorem subcand

Description: Cancellation law for subtraction. (Contributed by Mario Carneiro, 27-May-2016)

Ref Expression
Hypotheses negidd.1 ⊢ φ → A ∈ ℂ
pncand.2 ⊢ φ → B ∈ ℂ
subaddd.3 ⊢ φ → C ∈ ℂ
subcand.4 ⊢ φ → A − B = A − C
Assertion subcand ⊢ φ → 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 subcand.4 ⊢ φ → A − B = A − C
5 subcan ⊢ A ∈ ℂ ∧ B ∈ ℂ ∧ C ∈ ℂ → A − B = A − C ↔ B = C
6 1 2 3 5 syl3anc ⊢ φ → A − B = A − C ↔ B = C
7 4 6 mpbid ⊢ φ → B = C