Metamath Proof Explorer


Theorem pnncan

Description: Cancellation law for mixed addition and subtraction. (Contributed by NM, 30-Jun-2005) (Revised by Mario Carneiro, 27-May-2016)

Ref Expression
Assertion pnncan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐴𝐶 ) ) = ( 𝐵 + 𝐶 ) )

Proof

Step Hyp Ref Expression
1 simp1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐴 ∈ ℂ )
2 simp2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ )
3 1 2 addcld ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐵 ) ∈ ℂ )
4 simp3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
5 subsub ( ( ( 𝐴 + 𝐵 ) ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐴𝐶 ) ) = ( ( ( 𝐴 + 𝐵 ) − 𝐴 ) + 𝐶 ) )
6 3 1 4 5 syl3anc ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐴𝐶 ) ) = ( ( ( 𝐴 + 𝐵 ) − 𝐴 ) + 𝐶 ) )
7 pncan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − 𝐴 ) = 𝐵 )
8 7 3adant3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − 𝐴 ) = 𝐵 )
9 8 oveq1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 + 𝐵 ) − 𝐴 ) + 𝐶 ) = ( 𝐵 + 𝐶 ) )
10 6 9 eqtrd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − ( 𝐴𝐶 ) ) = ( 𝐵 + 𝐶 ) )