Proof of Identity for Laplacian

Theorem
\[ \mathbf{\nabla} \cdot (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \nabla^2 f \]
  where
\[g\]
  is an arbitrary function.
Proof
\[\mathbf{\nabla} \cdot (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \mathbf{\nabla} (\mathbf{\nabla} \times g + \mathbf{\nabla} f = \mathbf{\nabla} \cdot ( \mathbf{\nabla} \times g) + \mathbf{\nabla} \cdot ( \mathbf{\nabla} f) = \mathbf{\nabla} \cdot ( \mathbf{\nabla} f) = \nabla^2 f\]

since  
\[\mathbf{\nabla} \cdot ( \mathbf{\nabla} \times g)=0 \]
  if
\[g\]
  is twice differentiable.

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