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Proof step 5 from proof Sep-eq (previous step)

((adP ⊗_ st))(H), ∀ a. a ∈ st → adP(a, st[a]) = adP0(a, st[a]) ⊦ ((adP0 ⊗_ st))(H)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
∀ a, H. a ∈ st ∧ adP0(a, st[a])(H) → adP(a, st[a])(H) ⊦ ((adP0 ⊗_ st))(H) → ((adP ⊗_ st))(H)
Proof step 7:
((adP ⊗_ st))(H), ∀ a. a ∈ st → adP(a, st[a]) = adP0(a, st[a]) ⊦ ((adP0 ⊗_ st))(H), (∀ a, H. a ∈ st ∧ adP(a, st[a])(H) → adP0(a, st[a])(H)) ∧ ((adP ⊗_ st))(H)

Rule argument

spec: <toplevel> name: Sep-forall seq: ∀ a, H. a ∈ st ∧ adP0(a, st[a])(H) → adP(a, st[a])(H) ⊦ ((adP0 ⊗_ st))(H) → ((adP ⊗_ st))(H) subst: [adP0, H, st, adP] → [adP, H, st, adP0] flags: <all paths>

Additional information

Used terms: adP, H, st, adP0