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Proof step 1 from proof sub-in (conclusion)

⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s0

Rule argument

Additional information

Used simplifier rules: Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2);%ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);