[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 15 from proof subarray-split-at (previous step)

# ar = n0 + n, n1 < n0 + n, n ≤ n1 ⊦ ar[n1] = (subarray(ar, 0, n) + subarray(ar, n, n0))[n1], n0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0]
Proof step 17:
# ar = n0 + n, n1 < n0 + n, n1 < # subarray(ar, 0, n) + # subarray(ar, n, n0), n ≤ n1, # subarray(ar, 0, n) ≤ n1 ⊦ ar[n1] = subarray(ar, n, n0)[n1 - # subarray(ar, 0, n)], n0 = 0

Rule argument

spec: <toplevel> name: append-at2 seq: ⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0] subst: [n, ar1, ar0] → [n1, subarray(ar, n, n0), subarray(ar, 0, n)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ # subarray(ar, m, n) = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1, subarray(ar, n, n0), subarray(ar, 0, n)