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Proof step 5 from proof Sep-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)

User applied rule cut formula

And obtained the following new goals

Proof step 6:
((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)

Rule argument

fma: s = ∅

Additional information

Used simplifier rules: Sep-emp: ⊦ aP ⊗_ ∅ = emp;% ⊦ emp(H) ↔ H = ∅; (s) Used formulas: s = ∅