Metamath Proof Explorer


Theorem npncan2

Description: Cancellation law for subtraction. (Contributed by Scott Fenton, 21-Jun-2013)

Ref Expression
Assertion npncan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐴 ) ) = 0 )

Proof

Step Hyp Ref Expression
1 npncan ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐴 ) ) = ( 𝐴 − 𝐴 ) )
2 1 3anidm13 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐴 ) ) = ( 𝐴 − 𝐴 ) )
3 subid ⊢ ( 𝐴 ∈ ℂ → ( 𝐴 − 𝐴 ) = 0 )
4 3 adantr ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐴 ) = 0 )
5 2 4 eqtrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 − 𝐵 ) + ( 𝐵 − 𝐴 ) ) = 0 )