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Proof step 7 from proof at-fill-h0 (previous step)

n < m0, n < # ar, ∀ n0, ar0, a0. n0 < m0 ∧ n0 < # ar0 → fill(ar0, a0, 0, m0)[n0] = a0 ∨ # ar0 = 0 ⊦ m0 = 0, n = m0, fill(ar, a, 0, m0)[n] = a, # ar = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n, ar, a

Additional information

Used terms: n, ar, a