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Proof step 1 from proof at-fill-other (conclusion)

⊦ n ≥ m ∧ n < # ar → fill(ar, a, 0, m)[n] = ar[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < # ar, m ≤ n, m < # ar ⊦ fill(ar, a, 0, m)[n] = ar[n], # ar = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%ge: ⊦ n ≥ m ↔ m ≤ n; from specification nat