Metamath Proof Explorer


Theorem pncan3

Description: Subtraction and addition of equals. (Contributed by NM, 14-Mar-2005) (Proof shortened by Steven Nguyen, 8-Jan-2023)

Ref Expression
Assertion pncan3 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + ( 𝐵𝐴 ) ) = 𝐵 )

Proof

Step Hyp Ref Expression
1 subcl ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( 𝐵𝐴 ) ∈ ℂ )
2 eqid ( 𝐵𝐴 ) = ( 𝐵𝐴 )
3 subadd ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ ( 𝐵𝐴 ) ∈ ℂ ) → ( ( 𝐵𝐴 ) = ( 𝐵𝐴 ) ↔ ( 𝐴 + ( 𝐵𝐴 ) ) = 𝐵 ) )
4 2 3 mpbii ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ ( 𝐵𝐴 ) ∈ ℂ ) → ( 𝐴 + ( 𝐵𝐴 ) ) = 𝐵 )
5 1 4 mpd3an3 ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( 𝐴 + ( 𝐵𝐴 ) ) = 𝐵 )
6 5 ancoms ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 + ( 𝐵𝐴 ) ) = 𝐵 )