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Proof step 11 from proof at-fill-h (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n, ar, a, m0 abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s) Used terms: n, ar, a, m0