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

⊦ n < # ar → fill(ar, a, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ a;ar[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < # ar ⊦ fill(ar, a, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ a;ar[n]), # ar = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1