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Proof step 6 from proof ls-unfold1 (previous step)

∃ r0. (((r => a0 × r0 * ls(r0, x0)) * P))(H) ⊦ r = null, ∃ a, y, r0. ((r => a × r0 * ls(r0, y) * P))(H) ∧ a0 ' + x0 = a ' + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information