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Proof step 6 from proof Sep-any (previous step)

(((adP(a, d) * adP ⊗_ st0) * P))(H) ⊦ a ∈ st0, ∃ a0, d0, st1. ¬ a0 ∈ st1 ∧ st0[a, d] = st1[a0, d0] ∧ ((adP(a0, d0) * adP ⊗_ st1 * P))(H)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: a, d, st0

Additional information

Used simplifier rules: sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep Used terms: a, d, st0