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Metamath Proof Explorer

Theorem sqcld

Description: Closure of square. (Contributed by Mario Carneiro, 28-May-2016)

Ref Expression
Hypothesis expcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
Assertion sqcld ( 𝜑 → ( 𝐴 ↑ 2 ) ∈ ℂ )

Proof

Step Hyp Ref Expression
1 expcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 sqcl ⊢ ( 𝐴 ∈ ℂ → ( 𝐴 ↑ 2 ) ∈ ℂ )
3 1 2 syl ⊢ ( 𝜑 → ( 𝐴 ↑ 2 ) ∈ ℂ )