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Proof step 24 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n3 < n2 + n1, n0 < # ar, n0 + n1 ≤ # ar ⊦ (n3 < n1 ⊃ a;ar[n3 + n0]) = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n3]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]
Proof step 26:
# ar = n2 + n0 + n1, n0 < # ar, n3 < n2 + n1, n0 + n1 ≤ # ar ⊦ (n3 < n1 ⊃ a;ar[n3 + n0]) = (fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n3], n3 < # fill(mkarray(n1), a)

Rule argument

spec: <toplevel> name: append-at1 seq: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n] subst: [n, ar1, ar0] → [n3, subarray(ar, n0 + n1, n2), fill(mkarray(n1), a)] flags: <all paths>

Additional information

Used simplifier rules: ⊦ # fill(ar, a) = # ar; (s)% ⊦ # mkarray(n) = n; (s)% ⊦ m < n → fill(mkarray(n), a)[m] = a; (s) Used terms: n3, subarray(ar, n0 + n1, n2), fill(mkarray(n1), a)