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Proof step 6 from proof ext (previous step)

a ∈ s1, ∀ a. a ∈ s1 ↔ a ∈ s2 ⊦ a ∈ s2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ∈ s1, ∀ a. a ∈ s1 ↔ a ∈ s2 ⊦ s1 = ∅, a ∈ s2

Rule argument

Additional information

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