Metamath Proof Explorer


Theorem sqsubswap

Description: Swap the order of subtraction in a square. (Contributed by Scott Fenton, 10-Jun-2013)

Ref Expression
Assertion sqsubswap ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴𝐵 ) ↑ 2 ) = ( ( 𝐵𝐴 ) ↑ 2 ) )

Proof

Step Hyp Ref Expression
1 subcl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴𝐵 ) ∈ ℂ )
2 sqneg ( ( 𝐴𝐵 ) ∈ ℂ → ( - ( 𝐴𝐵 ) ↑ 2 ) = ( ( 𝐴𝐵 ) ↑ 2 ) )
3 1 2 syl ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( - ( 𝐴𝐵 ) ↑ 2 ) = ( ( 𝐴𝐵 ) ↑ 2 ) )
4 negsubdi2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → - ( 𝐴𝐵 ) = ( 𝐵𝐴 ) )
5 4 oveq1d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( - ( 𝐴𝐵 ) ↑ 2 ) = ( ( 𝐵𝐴 ) ↑ 2 ) )
6 3 5 eqtr3d ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴𝐵 ) ↑ 2 ) = ( ( 𝐵𝐴 ) ↑ 2 ) )