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Proof step 2 from proof le-one-01 (previous step)

# x + # y ≤ 1 ⊦ x = [], # x = 1 ∧ y = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# x + # y ≤ 1 ⊦ x = [], x.tail = [] ∧ y = []

Rule argument

Additional information

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