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Proof step 2 from proof get-finto (previous step)

m < n ⊦ (x to n)[m] = x[m], # x < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < n, m < # x, n ≤ # x ⊦ (x to n)[m] = x[m], x = []

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)