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Proof step 6 from proof sharing-emp (previous step)

∃ H0, H1, H2. H0 ⊥ H2 ∧ H0 ⊥ H1 ∧ H1 ⊥ H2 ∧ H = H0 ∪ H1 ∪ H2 ∧ P(H0 ∪ H2) ∧ H1 ∪ H2 = ∅ ⊦ P(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
H1 ∪ H0 = ∅, P(H2 ∪ H0), H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ H0 ⊦ P(H2)

Rule argument

Additional information

Used simplifier rules: ⊦ H ∪ ∅ = H; (s)