# Metamath Proof Explorer

## Theorem ppncan

Description: Cancellation law for mixed addition and subtraction. (Contributed by NM, 30-Jun-2005)

Ref Expression
Assertion ppncan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) + ( 𝐶𝐵 ) ) = ( 𝐴 + 𝐶 ) )

### Proof

Step Hyp Ref Expression
1 addcom ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + 𝐵 ) = ( 𝐵 + 𝐴 ) )
2 1 3adant3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐵 ) = ( 𝐵 + 𝐴 ) )
3 2 oveq1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐵𝐶 ) ) = ( ( 𝐵 + 𝐴 ) − ( 𝐵𝐶 ) ) )
4 addcl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + 𝐵 ) ∈ ℂ )
5 4 3adant3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐵 ) ∈ ℂ )
6 subsub2 ( ( ( 𝐴 + 𝐵 ) ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐵𝐶 ) ) = ( ( 𝐴 + 𝐵 ) + ( 𝐶𝐵 ) ) )
7 5 6 syld3an1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐵𝐶 ) ) = ( ( 𝐴 + 𝐵 ) + ( 𝐶𝐵 ) ) )
8 pnncan ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) − ( 𝐵𝐶 ) ) = ( 𝐴 + 𝐶 ) )
9 8 3com12 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) − ( 𝐵𝐶 ) ) = ( 𝐴 + 𝐶 ) )
10 3 7 9 3eqtr3d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) + ( 𝐶𝐵 ) ) = ( 𝐴 + 𝐶 ) )