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Proof step 1 from proof in-09 (conclusion)

⊦ ¬ a ∈ x.rest → (a ∈ x ↔ a = x.first ∧ x ≠ [])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ x ⊦ a = x.first, x = [], a ∈ x.rest

Rule argument

Additional information

Used simplifier rules: in-08: ⊦ x.first ∈ x ↔ x ≠ [];%in-empty: ⊦ ¬ a ∈ [];