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Proof step 6 from proof Sep-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
aP(a)(H0), a ∈ s, ∀ a. a ∈ s → aP(a) = aP0(a) ⊦ # s = 0, ((aP ⊗_ s))(H) ↔ ((aP0 ⊗_ s))(H), s = ∅, aP0(a)(H0)

Rule argument

Additional information

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