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Proof step 10 from proof sharing-emp (previous step)

P(H2 ∪ H0), H1 ⊥ H0, H2 ⊥ H1, H2 ⊥ H0, ∀ r. (r ∈ H1 ∪ H0 ↔ r ∈ ∅) ∧ (r ∈ H1 ∪ H0 → (H1 ∪ H0)[r] = (∅)[r]) ⊦ P(H2), H1 = ∅ ∧ H0 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
P(H2 ∪ H0), H2 ⊥ H0, H2 ⊥ H1, H1 ⊥ H0, ∀ r. ¬ r ∈ H1 ∧ ¬ r ∈ H0 ⊦ P(H2), H1 = ∅ ∧ H0 = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ r ∈ ∅; (s)% ⊦ H0 ⊥ H1 → (r ∈ H0 ∪ H1 ↔ r ∈ H0 ∨ r ∈ H1); (s)