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Proof step 22 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]) = (ar0 + fill(ar1, a, n0, m))[n] |
User applied rule insert rewrite lemma
And obtained the following new goals
- Proof step 23:
⊦ n < # ar0 + # ar1 → (ar0 + ar1)[n] = (n < # ar0 ⊃ ar0[n];ar1[n - # ar0])
- Proof step 24:
n < # ar0 + # ar1, n < # ar0 + # fill(ar1, a, n0, m), # ar0 ≤ n
⊦ # ar1 = 0,
(n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;ar1[n - # ar0]) = (n < # ar0 ⊃ ar0[n];fill(ar1, a, n0, m)[n - # ar0])
Rule argument
spec: <toplevel>
name: append-at
seq: ⊦ n < # ar0 + # ar1
→ (ar0 + ar1)[n]
= (n < # ar0 ⊃ ar0[n];ar1[n - # ar0])
subst: [n, ar1, ar0] → [n, fill(ar1, a, n0, m), ar0]
flags: <all paths>Additional information
Used simplifier rules:
⊦ # fill(ar, a, m, n) = # ar; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)
Used terms: n, fill(ar1, a, n0, m), ar0