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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: Fill-succ: ⊦ fill(ar, a, m, n + 1) = fill(ar, a, m, n)[m + n, a];%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2