Metamath Proof Explorer


Theorem nnncan1d

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 )
Assertion nnncan1d ( 𝜑 → ( ( 𝐴𝐵 ) − ( 𝐴𝐶 ) ) = ( 𝐶𝐵 ) )

Proof

Step Hyp Ref Expression
1 negidd.1 ( 𝜑𝐴 ∈ ℂ )
2 pncand.2 ( 𝜑𝐵 ∈ ℂ )
3 subaddd.3 ( 𝜑𝐶 ∈ ℂ )
4 nnncan1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) − ( 𝐴𝐶 ) ) = ( 𝐶𝐵 ) )
5 1 2 3 4 syl3anc ( 𝜑 → ( ( 𝐴𝐵 ) − ( 𝐴𝐶 ) ) = ( 𝐶𝐵 ) )