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

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

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 ∨ m0 = 0 ∨ # ar = 0 ⊦ # ar = 0, 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];%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2