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Proof step 17 from proof fill-append-shift (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]
Proof step 19:
n < # ar0 + # ar1 ⊦ (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;(ar0 + ar1)[n]) = (ar0 + fill(ar1, a, n0, m))[n], n < # ar0
Proof step 29:
n < # ar0 + # ar1, n < # ar0 ⊦ (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;(ar0 + ar1)[n]) = ar0[n]

Rule argument

spec: <toplevel> name: append-at1 seq: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n] subst: [n, ar1, ar0] → [n, fill(ar1, a, n0, m), ar0] flags: <all paths>

Additional information

Used terms: n, fill(ar1, a, n0, m), ar0