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Proof step 15 from proof at-fill-h (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n, ar, a, m0

Additional information

Used terms: n, ar, a, m0