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Proof step 13 from proof Sep-forall (previous step)

((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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 15:
∀ a0, H0. (a0 = a ∨ a0 ∈ s0) ∧ aP0(a0)(H0) → aP(a0)(H0), ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ aP0(a)(H0) ∧ ((aP0 ⊗_ s0))(H1), ∀ 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

spec: heap-sep name: sep seq: ⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)) subst: [H, P1, P0] → [H, aP0 ⊗_ s0, aP0(a)] flags: <all paths>

Additional information

Used terms: H, aP0 ⊗_ s0, aP0(a)