Metamath Proof Explorer


Theorem nnncan2

Description: Cancellation law for subtraction. (Contributed by NM, 1-Oct-2005)

Ref Expression
Assertion nnncan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐶 ) − ( 𝐵𝐶 ) ) = ( 𝐴𝐵 ) )

Proof

Step Hyp Ref Expression
1 subcl ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵𝐶 ) ∈ ℂ )
2 1 3adant1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵𝐶 ) ∈ ℂ )
3 sub32 ( ( 𝐴 ∈ ℂ ∧ ( 𝐵𝐶 ) ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) − 𝐶 ) = ( ( 𝐴𝐶 ) − ( 𝐵𝐶 ) ) )
4 2 3 syld3an2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) − 𝐶 ) = ( ( 𝐴𝐶 ) − ( 𝐵𝐶 ) ) )
5 nnncan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 − ( 𝐵𝐶 ) ) − 𝐶 ) = ( 𝐴𝐵 ) )
6 4 5 eqtr3d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴𝐶 ) − ( 𝐵𝐶 ) ) = ( 𝐴𝐵 ) )