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Proof step 8 from proof lst-in-hp (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
rs1 ∩ rs0 = ∅, o[ls(r, x)][rs1, cf, af], R[rs0, cf, af] ⊦ r = null, r ∈ rs1, r ∈ rs0

Rule argument

Additional information

Used simplifier rules: unc: ⊦ rs ∪ rs0 = rs0 ∪ rs; from specification set-union[refset]%una: ⊦ (rs ∪ rs0) ∪ rs1 = rs ∪ rs0 ∪ rs1; from specification set-union[refset]% ⊦ r ∈ rs ∪ rs0 ↔ ¬(¬ r ∈ rs ∧ ¬ r ∈ rs0); (ss)