Metamath Proof Explorer


Theorem nppcan3

Description: Cancellation law for subtraction. (Contributed by Mario Carneiro, 14-Sep-2015)

Ref Expression
Assertion nppcan3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) + ( 𝐶 + 𝐵 ) ) = ( 𝐴 + 𝐶 ) )

Proof

Step Hyp Ref Expression
1 subcl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴𝐵 ) ∈ ℂ )
2 1 3adant3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴𝐵 ) ∈ ℂ )
3 simp3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐶 ∈ ℂ )
4 simp2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → 𝐵 ∈ ℂ )
5 2 3 4 addassd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴𝐵 ) + 𝐶 ) + 𝐵 ) = ( ( 𝐴𝐵 ) + ( 𝐶 + 𝐵 ) ) )
6 nppcan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴𝐵 ) + 𝐶 ) + 𝐵 ) = ( 𝐴 + 𝐶 ) )
7 5 6 eqtr3d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐵 ) + ( 𝐶 + 𝐵 ) ) = ( 𝐴 + 𝐶 ) )