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Proof step 12 from proof range-eq-splice-less (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ m ≠ 0 → # splice(ar0, n0, ar1, n1, m) = max(# ar1, n1 + m)
Proof step 14:
⊦ splice(ar0, n0, ar1, n1, 0) = ar1
Proof step 15:
n3 < m0, n3 < m1, # ar0 < n0 + m1, m1 ≤ m0, n + m0 ≤ # ar, n1 ≤ n0, n2 + m1 ≤ # ar2, n1 + m1 ≤ max(# ar0, n0 + m0), ∀ n3. n3 < m1 → ar2[n2 + n3] = splice(ar, n, ar0, n0, m0)[n1 + n3] ⊦ ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m1), n0, m1)[n1 + n3], m0 = 0

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: splice-0: ⊦ splice(ar0, n0, ar1, n1, 0) = ar1;% ⊦ 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)% ⊦ m < n → n ≠ 0; (s) Used terms: m0, n0, ar0, n, ar