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Proof step 5 from proof at-fill-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n < # ar, ∀ n, ar, a, m0. n < # ar → fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ n < m0 + m ⊃ a;ar[n]) ∨ # ar = 0 ⊦ n = m0 + m, # ar = 0, fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ ¬ m0 + m < n ⊃ a;ar[n])

Rule argument

Additional information

Used simplifier rules: Put-other: ⊦ m ≠ n → ar[n, a][m] = ar[m]; from specification array