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Proof step 6 from proof sep-partial-mp-2 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
((P0 * (P0 * P2 -* P1)))(H1), ((P0 * P * (P0 * P2 -* P1)))(H0 ∪ H1), P(H0), H0 ⊥ H1, ∀ P0. ((P * P0))(H0 ∪ H1) → (∃ H2, H3. H2 ⊥ H3 ∧ H2 ∪ H3 = H0 ∪ H1 ∧ P(H2) ∧ P0(H3)) ⊦ ((P * (P2 -* P1)))(H0 ∪ H1)

Rule argument

Additional information

Used simplifier rules: sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2;%sep-com: ⊦ P0 * P1 = P1 * P0;