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Proof step 5 from proof owner-ls1 (previous step)

∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ o[ls(r, x)][rs0, cf, af] ∧ R[rs1, cf, af] ⊦ af(r) = o, r = null

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information