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Proof step 9 from proof Sep-eq (previous step)

((aP0 ⊗_ s))(H), ∀ a. a ∈ s → aP(a) = aP0(a) ⊦ ((aP ⊗_ s))(H)

User applied rule apply lemma

And obtained the following new goals

Proof step 10:
∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H) ⊦ ((aP0 ⊗_ s))(H) → ((aP ⊗_ s))(H)
Proof step 11:
((aP0 ⊗_ s))(H), ∀ a. a ∈ s → aP(a) = aP0(a) ⊦ ((aP ⊗_ s))(H), ∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H)

Rule argument

spec: <toplevel> name: Sep-forall seq: ∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H) ⊦ ((aP0 ⊗_ s))(H) → ((aP ⊗_ s))(H) subst: [s, aP0, aP, H] → [s, aP0, aP, H]

Additional information

Used terms: s, aP0, aP, H