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Proof step 9 from proof fill2append (previous step)

# ar = n2 + n0 + n1, n < # ar, n0 + n1 ≤ # ar ⊦ fill(ar, a, n0, n1)[n] = (subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ n < # ar → fill(ar, a, m0, m1)[n] = (m0 ≤ n ∧ n < m0 + m1 ⊃ a;ar[n])
Proof step 11:
# ar = n2 + n0 + n1, n < # ar, n0 + n1 ≤ # ar ⊦ (n0 ≤ n ∧ n < n0 + n1 ⊃ a;ar[n]) = (subarray(ar, 0, n0) + fill(mkarray(n1), a) + subarray(ar, n0 + n1, n2))[n]

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, n1, n0, a, ar] flags: <all paths>

Additional information

Used terms: n, n1, n0, a, ar