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Proof step 2 from proof at-fill-other (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < # ar, n < # ar, m ≤ n ⊦ fill(ar, a, 0, m)[n] = ar[n]

Rule argument

Additional information

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