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Proof step 2 from proof in-09 (previous step)

⊦ a ∈ x ↔ a = x.head ∧ x ≠ [], a ∈ x.tail

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ x ⊦ a = x.head, x = [], a ∈ x.tail

Rule argument

Additional information

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