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Proof step 18 from proof Sep-forall (previous step)

(adP0 ⊗_ st)(H) ∧ (∀ a, H. a ∈ st ∧ adP0(a, st[a])(H) → adP(a, st[a])(H)), ∀ st0, adP0, H, adP. # st0 < # st → ((adP0 ⊗_ st0)(H) ∧ (∀ a, H. a ∈ st0 ∧ adP0(a, st0[a])(H) → adP(a, st0[a])(H)) → (adP ⊗_ st0)(H)) ⊦ (adP ⊗_ st)(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
(adP0 ⊗_ st)(H), ∀ a, H. a ∈ st ∧ adP0(a, st[a])(H) → adP(a, st[a])(H), ∀ st0, adP0, H, adP. # st0 < # st → ((adP0 ⊗_ st0)(H) ∧ (∀ a, H. a ∈ st0 ∧ adP0(a, st0[a])(H) → adP(a, st0[a])(H)) → (adP ⊗_ st0)(H)) ⊦ (adP ⊗_ st)(H)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]