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Proof step 1 from proof And-forall (conclusion)

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

System applied rule normalize

And obtained the following new goals

Proof step 2:
((aP0 ∧_ s))(H), ∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H) ⊦ ((aP ∧_ s))(H)

Rule argument

Additional information