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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1)
Proof step 18:
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(a)(H1) ∧ ((aP ⊗_ s0))(H0) ∧ H1 ⊥ H0

Rule argument

spec: heap-sep name: split-union seq: ⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1) subst: [H1, H0, P1, P0] → [H0, H1, aP ⊗_ s0, aP(a)] flags: <all paths>

Additional information

Used terms: H0, H1, aP ⊗_ s0, aP(a)