Metamath Proof Explorer


Theorem subcl

Description: Closure law for subtraction. (Contributed by NM, 10-May-1999) (Revised by Mario Carneiro, 21-Dec-2013)

Ref Expression
Assertion subcl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴𝐵 ) ∈ ℂ )

Proof

Step Hyp Ref Expression
1 subval ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴𝐵 ) = ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) )
2 negeu ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ∃! 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 )
3 2 ancoms ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ∃! 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 )
4 riotacl ( ∃! 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 → ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) ∈ ℂ )
5 3 4 syl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) ∈ ℂ )
6 1 5 eqeltrd ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴𝐵 ) ∈ ℂ )