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Metamath Proof Explorer

Theorem sqmuld

Description: Distribution of squaring over multiplication. (Contributed by Mario Carneiro, 28-May-2016)

Ref Expression
Hypotheses expcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
mulexpd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
Assertion sqmuld ( 𝜑 → ( ( 𝐴 · 𝐵 ) ↑ 2 ) = ( ( 𝐴 ↑ 2 ) · ( 𝐵 ↑ 2 ) ) )

Proof

Step Hyp Ref Expression
1 expcld.1 ⊢ ( 𝜑 → 𝐴 ∈ ℂ )
2 mulexpd.2 ⊢ ( 𝜑 → 𝐵 ∈ ℂ )
3 sqmul ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 · 𝐵 ) ↑ 2 ) = ( ( 𝐴 ↑ 2 ) · ( 𝐵 ↑ 2 ) ) )
4 1 2 3 syl2anc ⊢ ( 𝜑 → ( ( 𝐴 · 𝐵 ) ↑ 2 ) = ( ( 𝐴 ↑ 2 ) · ( 𝐵 ↑ 2 ) ) )