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Proof step 25 from proof fill-append-shift (previous step)

n < # ar0 + # ar1, # ar0 ≤ n ⊦ # ar1 = 0, (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;ar1[n - # ar0]) = fill(ar1, a, n0, m)[n - # ar0]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 26:
n = n1 + # ar0, n < # ar0 + # ar1, # ar0 ≤ n ⊦ # ar1 = 0, (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;ar1[n1]) = fill(ar1, a, n0, m)[n1]

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, # ar0, n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # ar0, n1