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Proof step 1 from proof fr-01 (conclusion)

⊦ x ≠ [] ∧ x ≠ x.first + [] → x.first + x.rest.first + x.rest.rest = x

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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