Metamath Proof Explorer


Theorem pnpcan2

Description: Cancellation law for mixed addition and subtraction. (Contributed by Scott Fenton, 9-Jun-2006)

Ref Expression
Assertion pnpcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐶 ) ) = ( 𝐴𝐵 ) )

Proof

Step Hyp Ref Expression
1 addcom ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐶 ) = ( 𝐶 + 𝐴 ) )
2 1 3adant2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐶 ) = ( 𝐶 + 𝐴 ) )
3 addcom ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) )
4 3 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) )
5 2 4 oveq12d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐶 ) ) = ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) )
6 pnpcan ( ( 𝐶 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) = ( 𝐴𝐵 ) )
7 6 3coml ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐶 + 𝐴 ) − ( 𝐶 + 𝐵 ) ) = ( 𝐴𝐵 ) )
8 5 7 eqtrd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐵 + 𝐶 ) ) = ( 𝐴𝐵 ) )