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Proof step 2 from proof intexin (previous step)

⊦ s ∩ s0 = ∅, ∃ a. a ∈ s ∧ a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s ⊦ ∃ a. a ∈ s ∧ a ∈ s0

Rule argument

Additional information

Used simplifier rules: Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification infset-basic% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)