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Proof step 5 from proof sep (previous step)

∃ H0, H1. H0 ⊥ H1 ∧ mk(rs, f) = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1) ⊦ ∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ P0(mk(rs0, f)) ∧ P1(mk(rs1, f))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information