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Proof step 2 from proof fr-01 (previous step)

⊦ x.head + x.tail.head + x.tail.tail = x, x = x.head + [], x = []

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: fr: ⊦ x ≠ [] ∧ x ≠ x.head + [] → x.tail.head + x.tail.tail = x.tail;%fr-02: ⊦ x ≠ [] → x.head + x.tail = x;