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Proof step 2 from proof fr-04 (previous step)

x.rest = y.rest, x.first = y.first ⊦ x = y, x = [], y = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [x, a, y0] flags: rewrite, current

Additional information

Used simplifier rules: fr-02: ⊦ x ≠ [] → x.first + x.rest = x; Used terms: x, a, y0