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Proof step 7 from proof at-fill-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n < # ar, m < # ar, ∀ n, ar, a, m0. n < # ar → fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ n < m0 + m ⊃ a;ar[n]) ⊦ fill(ar, a, m0, m)[n] = (m0 ≤ n ∧ ¬ m0 + m < n ⊃ a;ar[n]), n = m0 + m

Rule argument

Additional information

Used simplifier rules: ⊦ # ar[n, a] = # ar; (s)