Metamath Proof Explorer


Theorem sqcl

Description: Closure of square. (Contributed by NM, 10-Aug-1999)

Ref Expression
Assertion sqcl ( 𝐴 ∈ ℂ → ( 𝐴 ↑ 2 ) ∈ ℂ )

Proof

Step Hyp Ref Expression
1 sqval ( 𝐴 ∈ ℂ → ( 𝐴 ↑ 2 ) = ( 𝐴 · 𝐴 ) )
2 mulcl ( ( 𝐴 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( 𝐴 · 𝐴 ) ∈ ℂ )
3 2 anidms ( 𝐴 ∈ ℂ → ( 𝐴 · 𝐴 ) ∈ ℂ )
4 1 3 eqeltrd ( 𝐴 ∈ ℂ → ( 𝐴 ↑ 2 ) ∈ ℂ )