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Proof step 4 from proof at-fill-other (previous step)

m0 < # ar, m0 < n, n < # ar, m0 + 1 < # ar, ∀ n, ar, a. n < # ar ∧ m0 ≤ n ∧ m0 < # ar → fill(ar, a, 0, m0)[n] = ar[n] ∨ # ar = 0 ⊦ n = 0, fill(ar, a, 0, m0)[n] = ar[n], # ar = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n, ar, a

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat Used terms: n, ar, a