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Proof step 14 from proof at-fill-h (previous step)

n < m, n < # ar, # ar ≤ m, ∀ 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, m0 + m < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
n < m0 + m, n < m, n < # ar, # ar ≤ m, ∀ 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 ⊃ a;ar[n]), n = m0 + m

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss)