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Proof step 6 from proof Imp2 (previous step)

P(H1), P0(H0), H1 ⊥ H0, ∀ H. P0(H) → P1(H) ⊦ ((P * P1))(H1 ∪ H0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: H0

Additional information

Used simplifier rules: split-union: ⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1);%sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2;%sep-com: ⊦ P0 * P1 = P1 * P0; Used terms: H0