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Proof step 6 from proof at-fill-h0 (previous step)

n < # ar, n ≤ m0, ∀ n, ar, a. n < m0 ∧ n < # ar → fill(ar, a, 0, m0)[n] = a ⊦ fill(ar, a, 0, m0)[n] = a, n = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n < m0, n < # ar, ∀ n, ar, a. n < m0 ∧ n < # ar → fill(ar, a, 0, m0)[n] = a ⊦ fill(ar, a, 0, m0)[n] = a, n = m0

Rule argument

Additional information

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