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Proof step 1 from proof le-one-01 (conclusion)

⊦ # x + # y ≤ 1 ∧ x ≠ [] → # x = 1 ∧ y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# x + # y ≤ 1 ⊦ x = [], x.rest = [] ∧ y = []

Rule argument

Additional information

Used simplifier rules: lf-01: ⊦ # x = 1 ↔ x = x.first + [];%fr-07: ⊦ x ≠ [] → (x = x.first + y ↔ x.rest = y);