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Metamath Proof Explorer

Theorem nnncan1

Description: Cancellation law for subtraction. (Contributed by NM, 8-Feb-2005) (Proof shortened by Andrew Salmon, 19-Nov-2011)

Ref Expression
Assertion nnncan1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) − ( 𝐴 − 𝐶 ) ) = ( 𝐶 − 𝐵 ) )

Proof

Step Hyp Ref Expression
1 subcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐶 ) ∈ ℂ )
2 1 3adant2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐶 ) ∈ ℂ )
3 sub32 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ ( 𝐴 − 𝐶 ) ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) − ( 𝐴 − 𝐶 ) ) = ( ( 𝐴 − ( 𝐴 − 𝐶 ) ) − 𝐵 ) )
4 2 3 syld3an3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) − ( 𝐴 − 𝐶 ) ) = ( ( 𝐴 − ( 𝐴 − 𝐶 ) ) − 𝐵 ) )
5 nncan ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( 𝐴 − 𝐶 ) ) = 𝐶 )
6 5 3adant2 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − ( 𝐴 − 𝐶 ) ) = 𝐶 )
7 6 oveq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐴 − 𝐶 ) ) − 𝐵 ) = ( 𝐶 − 𝐵 ) )
8 4 7 eqtrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) − ( 𝐴 − 𝐶 ) ) = ( 𝐶 − 𝐵 ) )