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Proof step 3 from proof to-from-01 (previous step)

⌜n⌝ < # ⌜x⌝ ⊦ (⌜x⌝ to n) from n = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n < # x ⊦ (x to n) from n = [], x = []

Rule argument

Additional information

Used simplifier rules: to-fin: ⊦ ⌜x⌝ to n = ⌜x to n⌝ ;%from-fin: ⊦ ⌜x⌝ from n = ⌜x from n⌝ ;%fin-empty: ⊦ ⌜x⌝ = [] ↔ x = [];%size-fin: ⊦ # ⌜x⌝ = ⌜# x⌝ ;%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf