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Proof step 9 from proof at-fill-h (previous step)

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

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n, ar, a, m0

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: n, ar, a, m0