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Proof step 48 from proof range-eq-splice-less (previous step)

# ar0 < n0 + m1, n2 + m1 ≤ # ar2, n + m0 ≤ # ar, m1 ≤ m0, n1 ≤ n0, ∀ n3. n3 < m1 → ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m1), n0, m1)[n1 + n3] ⊦ m1 = 0, n1 + m1 ≤ # splice(ar, n, ar0, n0, m0) ∧ (n3 < m1 → ar2[n2 + n3] = splice(ar, n, ar0, n0, m0)[n1 + n3])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 49:
⊦ m ≠ 0 → # splice(ar0, n0, ar1, n1, m) = max(# ar1, n1 + m)
Proof step 50:
# ar0 < n0 + m1, n1 ≤ n0, m1 ≤ m0, n + m0 ≤ # ar, n2 + m1 ≤ # ar2, ∀ n3. n3 < m1 → ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m1), n0, m1)[n1 + n3] ⊦ m0 = 0, m1 = 0, n1 + m1 ≤ max(# ar0, n0 + m0) ∧ (n3 < m1 → ar2[n2 + n3] = splice(ar, n, ar0, n0, m0)[n1 + n3])

Rule argument

spec: <toplevel> name: splice-len seq: ⊦ m ≠ 0 → # splice(ar0, n0, ar1, n1, m) = max(# ar1, n1 + m) subst: [m, n1, ar1, n0, ar0] → [m0, n0, ar0, n, ar] flags: <all paths>

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ 0 ↔ m = 0; (s) Used terms: m0, n0, ar0, n, ar