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Proof step 1 from proof fr-05 (conclusion)

⊦ x ≠ [] → (x ≠ x.first + y ↔ x.rest ≠ y)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: fr-02: ⊦ x ≠ [] → x.first + x.rest = x;%fr-07: ⊦ x ≠ [] → (x = x.first + y ↔ x.rest = y);