Theorem nppcan

Description: Cancellation law for subtraction. (Contributed by NM, 1-Sep-2005)

	Ref	Expression
Assertion	nppcan	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐶 ) + 𝐵 ) = ( 𝐴 + 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		subcl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )
2	1	3adant3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ 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	add32d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐶 ) + 𝐵 ) = ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) + 𝐶 ) )
6		npcan	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + 𝐵 ) = 𝐴 )
7	6	oveq1d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) + 𝐶 ) = ( 𝐴 + 𝐶 ) )
8	7	3adant3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) + 𝐶 ) = ( 𝐴 + 𝐶 ) )
9	5 8	eqtrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐶 ) + 𝐵 ) = ( 𝐴 + 𝐶 ) )