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

∀ x. n0 ≤ # x → #(x to n0) = n0, n0 +1 ≤ # x ⊦ #(x to n0 +1) = n0 +1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 < # x, ∀ x. n0 ≤ # x → #(x to n0) = n0 ⊦ #(x to n0 + 1) = n0 + 1, x = []

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2