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

adP0(a, d)(H1), ((adP0 ⊗_ st0))(H0), H1 ⊥ H0, ∀ a0, H. a0 ∈ st0[a, d] ∧ adP0(a0, st0[a, d][a0])(H) → adP(a0, st0[a, d][a0])(H), ∀ st1, adP0, H, adP. # st1 < # st0[a, d] → ((adP0 ⊗_ st1))(H) ∧ (∀ a, H. a ∈ st1 ∧ adP0(a, st1[a])(H) → adP(a, st1[a])(H)) → ((adP ⊗_ st1))(H) ⊦ ((adP(a, d) * adP ⊗_ st0))(H1 ∪ H0), a ∈ st0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ P0(H0) ∧ P1(H1) ∧ H0 ⊥ H1 → ((P0 * P1))(H0 ∪ H1)
Proof step 14:
adP0(a, d)(H1), ((adP0 ⊗_ st0))(H0), H1 ⊥ H0, ∀ a0, H. a0 ∈ st0[a, d] ∧ adP0(a0, st0[a, d][a0])(H) → adP(a0, st0[a, d][a0])(H), ∀ st1, adP0, H, adP. # st1 < # st0[a, d] → ((adP0 ⊗_ st1))(H) ∧ (∀ a, H. a ∈ st1 ∧ adP0(a, st1[a])(H) → adP(a, st1[a])(H)) → ((adP ⊗_ st1))(H) ⊦ ((adP(a, d) * adP ⊗_ st0))(H1 ∪ H0), a ∈ st0, adP(a, d)(H1) ∧ ((adP ⊗_ st0))(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, adP ⊗_ st0, adP(a, d)] flags: <all paths>

Additional information

Used terms: H0, H1, adP ⊗_ st0, adP(a, d)