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Proof step 3 from proof at-fill-h0 (previous step)

n < # ar, n < m0 +1, ∀ n, ar, a. n < m0 ∧ n < # ar → fill(ar, a, 0, m0)[n] = a ⊦ fill(ar, a, 0, m0 +1)[n] = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n < # ar, n ≤ m0, ∀ n, ar, a. n < m0 ∧ n < # ar → fill(ar, a, 0, m0)[n] = a ⊦ fill(ar, a, 0, m0)[m0, a][n] = a

Rule argument

Additional information

Used simplifier rules: Fill-succ: ⊦ fill(ar, a, m, n + 1) = fill(ar, a, m, n)[m + n, a];% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ m +1 = m + 1; (s)