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Metamath Proof Explorer

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 ) )