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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 < # ar, m0 < n, n < # ar, m0 + 1 < # ar, ∀ n, ar, a. n < # ar ∧ m0 ≤ n ∧ m0 < # ar → fill(ar, a, 0, m0)[n] = ar[n] ∨ # ar = 0 ⊦ n = 0, fill(ar, a, 0, m0)[n] = ar[n], # ar = 0

Rule argument

Additional information

Used simplifier rules: Fill-succ: ⊦ fill(ar, a, m, n + 1) = fill(ar, a, m, n)[m + n, a];%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%cngetput: m ≠ n ⊦ ar[m, a][n] = ar[n]; from specification array%ref-1: ⊦ ¬ n < n; from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%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