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Proof step 12 from proof Sep-forall (previous step)

((aP0 ⊗_ s))(H), ∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H), ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0, ∀ s0, aP0, H, aP. # s0 < # s → ((aP0 ⊗_ s0))(H) ∧ (∀ a, H. a ∈ s0 ∧ aP0(a)(H) → aP(a)(H)) → ((aP ⊗_ s0))(H) ⊦ # s = 0, ((aP ⊗_ s))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
((aP0(a) * 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) ⊦ ((aP(a) * aP ⊗_ s0))(H), a ∈ s0

Rule argument

Additional information

Used simplifier rules: Sep-ins: ⊦ ¬ a ∈ s → aP ⊗_ (s ++ a) = aP(a) * aP ⊗_ s;% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1; (s)