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Proof step 64 from proof splice-resize-less (previous step)

n < max(# ar1, n1 + m0), n < n2 + n1, # ar1 < n1 + n2, n2 ≤ m0 ⊦ splice(ar0, n0, ar1, n1, m0)[n] = copy(ar0, n0, resize(ar1, n1 + n2), n1, n2)[n], m0 = 0, n2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 65:
⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0])
Proof step 66:
⊦ # resize(ar, n) = n
Proof step 67:
# ar1 < n1 + n2, n < n2 + n1, n < max(# ar1, n1 + m0), n < # resize(ar1, n1 + n2), n2 ≤ m0 ⊦ splice(ar0, n0, ar1, n1, m0)[n] = (n < n1 ∨ n1 + n2 ≤ n ⊃ resize(ar1, n1 + n2)[n];ar0[n + n0 - n1]), m0 = 0, n2 = 0

Rule argument

spec: array-copy name: at-copy seq: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]) subst: [m0, m, n0, ar1, n, ar0] → [n, n2, n1, resize(ar1, n1 + n2), n0, ar0] flags: <all paths>

Additional information

Used simplifier rules: resize-len: ⊦ # resize(ar, n) = n;% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: n, n2, n1, resize(ar1, n1 + n2), n0, ar0