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Proof step 16 from proof And-forall (previous step)

aP(a)(H), aP0(a)(H), ((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) ⊦ a ∈ s0, ((aP ∧_ s0))(H), ((aP0 ∧_ s0))(H) ∧ (∀ a, H. a ∈ s0 ∧ aP0(a)(H) → aP(a)(H))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
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)

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic