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Proof step 10 from proof wand-flatten (previous step)

H0 ⊥ H, ∀ H1. H1 ⊥ H ∧ P0(H1) → ((P1 -* P2))(H1 ∪ H), ∃ H2, H1. H2 ⊥ H1 ∧ H0 = H2 ∪ H1 ∧ P0(H2) ∧ P1(H1) ⊦ P2(H0 ∪ H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
P0(H2), P1(H1), H2 ⊥ H1, H ⊥ H2, H ⊥ H1, ∀ H1. H1 ⊥ H ∧ P0(H1) → ((P1 -* P2))(H1 ∪ H) ⊦ P2((H2 ∪ H1) ∪ H)

Rule argument

Additional information

Used simplifier rules: ⊦ H1 ⊥ H2 → (H1 ∪ H2 ⊥ H0 ↔ H0 ⊥ H1 ∧ H0 ⊥ H2); (ws)