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Proof step 5 from proof subdef (previous step)

∀ a. a ∈ s1 → a ∈ s2 ⊦ ∀ a. a ∈ s1 → a ∈ s2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ s1, a0 ∈ s1, ∀ a. a ∈ s1 → a ∈ s2 ⊦ a ∈ s2

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic