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Proof step 6 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) ⊦ s = ∅, ((aP ∧_ s))(H)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
((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) ⊦ # s = 0, s = ∅, ((aP ∧_ s))(H)

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic