Theorem sqdivid

Description: The square of a nonzero complex number divided by itself equals that number. (Contributed by AV, 19-Jul-2021)

	Ref	Expression
Assertion	sqdivid	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( ( 𝐴 ↑ 2 ) / 𝐴 ) = 𝐴 )

Proof

Step	Hyp	Ref	Expression
1		sqval	⊢ ( 𝐴 ∈ ℂ → ( 𝐴 ↑ 2 ) = ( 𝐴 · 𝐴 ) )
2	1	adantr	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( 𝐴 ↑ 2 ) = ( 𝐴 · 𝐴 ) )
3	2	oveq1d	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( ( 𝐴 ↑ 2 ) / 𝐴 ) = ( ( 𝐴 · 𝐴 ) / 𝐴 ) )
4		divcan3	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( ( 𝐴 · 𝐴 ) / 𝐴 ) = 𝐴 )
5	4	3anidm12	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( ( 𝐴 · 𝐴 ) / 𝐴 ) = 𝐴 )
6	3 5	eqtrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐴 ≠ 0 ) → ( ( 𝐴 ↑ 2 ) / 𝐴 ) = 𝐴 )