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Metamath Proof Explorer

Theorem nnncan

Description: Cancellation law for subtraction. (Contributed by NM, 4-Sep-2005)

Ref Expression
Assertion nnncan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 − 𝐶 ) ) − 𝐶 ) = ( 𝐴 − 𝐵 ) )

Proof

Step Hyp Ref Expression
1 subcl ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 − 𝐶 ) ∈ ℂ )
2 1 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 − 𝐶 ) ∈ ℂ )
3 subsub4 ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 − 𝐶 ) ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 − 𝐶 ) ) − 𝐶 ) = ( 𝐴 − ( ( 𝐵 − 𝐶 ) + 𝐶 ) ) )
4 2 3 syld3an2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 − 𝐶 ) ) − 𝐶 ) = ( 𝐴 − ( ( 𝐵 − 𝐶 ) + 𝐶 ) ) )
5 npcan ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐵 − 𝐶 ) + 𝐶 ) = 𝐵 )
6 5 oveq2d ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( ( 𝐵 − 𝐶 ) + 𝐶 ) ) = ( 𝐴 − 𝐵 ) )
7 6 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( ( 𝐵 − 𝐶 ) + 𝐶 ) ) = ( 𝐴 − 𝐵 ) )
8 4 7 eqtrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵 − 𝐶 ) ) − 𝐶 ) = ( 𝐴 − 𝐵 ) )