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• If the uses precede the assignments, we actually have an anti-dependence, ie
  • $S_p ~ \overline{\delta}_< ~ S_q$ if, for each solution pair (I₁,I₂), I₁ > I₂

• If there are some solution pairs (I₁,I₂) with I₁ < I₂ and some with I₁ > I₂, we write $S_p ~ \delta_* ~ S_q$

• If, for all solution pairs (I₁,I₂), I₁ = I₂, there are dependences within an iteration of the loop, but there are no loop-carried dependences.