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Proof step 10 from proof fill-append-shift (previous step)

n < # ar0 + # ar1 ⊦ fill(ar0 + ar1, a, n0 + # ar0, m)[n] = (ar0 + fill(ar1, a, n0, m))[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: array-copy name: at-fill-h seq: ⊦ n < # ar → fill(ar, a, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ a;ar[n]) subst: [n, m1, m0, a, ar] → [n, m, n0 + # ar0, a, ar0 + ar1] flags: <all paths>

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: n, m, n0 + # ar0, a, ar0 + ar1