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Metamath Proof Explorer

Theorem nnpcan

Description: Cancellation law for subtraction: ((a-b)-c)+b = a-c holds for complex numbers a,b,c. (Contributed by Alexander van der Vekens, 24-Mar-2018)

		Ref	Expression
	Assertion	nnpcan	⊢ ( ( 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		addsub	⊢ ( ( ( 𝐴 − 𝐵 ) ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) − 𝐶 ) = ( ( ( 𝐴 − 𝐵 ) − 𝐶 ) + 𝐵 ) )
4	3	eqcomd	⊢ ( ( ( 𝐴 − 𝐵 ) ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( ( 𝐴 − 𝐵 ) − 𝐶 ) + 𝐵 ) = ( ( ( 𝐴 − 𝐵 ) + 𝐵 ) − 𝐶 ) )
5	2 4	syld3an1	⊢ ( ( 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	3adant3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + 𝐵 ) = 𝐴 )
8	7	oveq1d	⊢ ( ( 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 ) )