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Proof step 17 from proof And-forall (previous step)

aP(a)(H), aP0(a0)(H0), aP0(a)(H), a0 ∈ s0, ((aP0 ∧_ s0))(H), ∀ 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(a0)(H0), a ∈ s0, ((aP ∧_ s0))(H)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: a0, H0

Additional information

Used terms: a0, H0