Metamath Proof Explorer


Theorem pncan3oi

Description: Subtraction and addition of equals. Almost but not exactly the same as pncan3i and pncan , this order happens often when applying "operations to both sides" so create a theorem specifically for it. A deduction version of this is available as pncand . (Contributed by David A. Wheeler, 11-Oct-2018)

Ref Expression
Hypotheses pncan3oi.1 𝐴 ∈ ℂ
pncan3oi.2 𝐵 ∈ ℂ
Assertion pncan3oi ( ( 𝐴 + 𝐵 ) − 𝐵 ) = 𝐴

Proof

Step Hyp Ref Expression
1 pncan3oi.1 𝐴 ∈ ℂ
2 pncan3oi.2 𝐵 ∈ ℂ
3 pncan ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 𝐵 ) − 𝐵 ) = 𝐴 )
4 1 2 3 mp2an ( ( 𝐴 + 𝐵 ) − 𝐵 ) = 𝐴