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Proof step 2 from proof le-leq-one-empty-or-head-equal-last (previous step)

# x ≤ 1 ⊦ x.head = x.last, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# x = 1 ⊦ x.head = x.last, x = []

Rule argument

Additional information

Used simplifier rules: not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%le-one: ⊦ # x ≤ 1 ∧ x ≠ [] → # x = 1; (forward) from specification list-data