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Proof step 16 from proof And-forall (previous step)

(aP0 ∧_ s)(H) ∧ (∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H)), ∀ s0, aP0, H, aP. # s0 < # s → ((aP0 ∧_ s0)(H) ∧ (∀ a, H. a ∈ s0 ∧ aP0(a)(H) → aP(a)(H)) → (aP ∧_ s0)(H)) ⊦ (aP ∧_ s)(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
(aP0 ∧_ s)(H), ∀ a, H. a ∈ s ∧ aP0(a)(H) → aP(a)(H), ∀ s0, aP0, H, aP. # s0 < # s → ((aP0 ∧_ s0)(H) ∧ (∀ a, H. a ∈ s0 ∧ aP0(a)(H) → aP(a)(H)) → (aP ∧_ s0)(H)) ⊦ (aP ∧_ s)(H)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]