Metamath Proof Explorer


Theorem subcand

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

Ref Expression
Hypotheses negidd.1 ( 𝜑𝐴 ∈ ℂ )
pncand.2 ( 𝜑𝐵 ∈ ℂ )
subaddd.3 ( 𝜑𝐶 ∈ ℂ )
subcand.4 ( 𝜑 → ( 𝐴𝐵 ) = ( 𝐴𝐶 ) )
Assertion subcand ( 𝜑𝐵 = 𝐶 )

Proof

Step Hyp Ref Expression
1 negidd.1 ( 𝜑𝐴 ∈ ℂ )
2 pncand.2 ( 𝜑𝐵 ∈ ℂ )
3 subaddd.3 ( 𝜑𝐶 ∈ ℂ )
4 subcand.4 ( 𝜑 → ( 𝐴𝐵 ) = ( 𝐴𝐶 ) )
5 subcan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) = ( 𝐴𝐶 ) ↔ 𝐵 = 𝐶 ) )
6 1 2 3 5 syl3anc ( 𝜑 → ( ( 𝐴𝐵 ) = ( 𝐴𝐶 ) ↔ 𝐵 = 𝐶 ) )
7 4 6 mpbid ( 𝜑𝐵 = 𝐶 )