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Proof step 8 from proof ext (previous step)

a ∈ s2, ∀ a. a ∈ s1 ↔ a ∈ s2 ⊦ a ∈ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a ∈ s2, ∀ a. a ∈ s1 ↔ a ∈ s2 ⊦ s2 = ∅, a ∈ s1

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic