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Proof step 23 from proof Sep-forall (previous step)

aP(a)(H1), aP0(a0)(H), aP0(a)(H1), a0 ∈ s0, ((aP0 ⊗_ s0))(H0), H1 ⊥ H0, ∀ a0, H0. (a0 = a ∨ a0 ∈ s0) ∧ aP0(a0)(H0) → aP(a0)(H0), ∀ s1, aP0, H, aP. # s1 < #(s0 ++ a) → ((aP0 ⊗_ s1))(H) ∧ (∀ a, H. a ∈ s1 ∧ aP0(a)(H) → aP(a)(H)) → ((aP ⊗_ s1))(H) ⊦ # s0 = 0, s0 = ∅, ((aP(a) * aP ⊗_ s0))(H1 ∪ H0), aP(a0)(H), a ∈ s0, ((aP ⊗_ s0))(H0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: a0, H

Additional information

Used terms: a0, H