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Proof step 11 from proof Sep-forall (previous step)

∀ a0, H. a0 ∈ st0[a, d] ∧ adP0(a0, st0[a, d][a0])(H) → adP(a0, st0[a, d][a0])(H), ∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ adP0(a, d)(H0) ∧ ((adP0 ⊗_ st0))(H1), ∀ 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))(H), a ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
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

Proof step 11 from proof Sep-forall (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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