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Proof step 5 from proof subset-equal (previous step)

a ∈ s0 → a ∈ s, ∀ a. a ∈ s0 → a ∈ s, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s ↔ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ s, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s0 → a ∈ s ⊦ a ∈ s0

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);