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Proof step 6 from proof ex-p (previous step)

P(H0), xP(x)(H1), H1 ⊥ H0 ⊦ ∃ x. ((xP(x) * P))(H1 ∪ H0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: x

Additional information

Used simplifier rules: ⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1); (ss)%sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep Used terms: x