Metamath Proof Explorer


Theorem sqdivid

Description: The square of a nonzero number divided by itself yields the number itself. (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 )