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Proof step 20 from proof Sep-forall (previous step)

aP0(a)(H1), ((aP0 ⊗_ s0))(H0), H1 ⊥ H0, (a = a ∨ a ∈ s0) ∧ aP0(a)(H1) → aP(a)(H1), ∀ 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))(H1 ∪ H0), a ∈ s0, aP(a)(H1) ∧ ((aP ⊗_ s0))(H0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
aP(a)(H1), aP0(a)(H1), ((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) ⊦ ((aP(a) * aP ⊗_ s0))(H1 ∪ H0), a ∈ s0, ((aP ⊗_ s0))(H0)

Proof step 20 from proof Sep-forall (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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