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Proof step 2 from proof to-empty-fin (previous step)

n < # x ⊦ x to n = [] ↔ n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x to n = [], n < # x ⊦ n = 0, x = []

Rule argument

Additional information

Used simplifier rules: to-zero-l: ⊦ x to 0 = [];% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic