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 ) → ( ( 𝐴𝐵 ) − ( 𝐴𝐶 ) ) = ( 𝐶𝐵 ) )

Proof

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