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Proof step 24 from proof splice-resize-less (previous step)

n < max(# ar1, n1 + m0), n < n1 ⊦ splice(ar0, n0, ar1, n1, m0)[n] = (# ar1 ≤ n ⊃ zero;ar1[n]), m0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ # ar1 < n1 + m ∧ m ≠ 0 → splice(ar0, n0, ar1, n1, m) = copy(ar0, n0, resize(ar1, n1 + m), n1, m)
Proof step 26:
⊦ n1 + m ≤ # ar1 ∧ m ≠ 0 → splice(ar0, n0, ar1, n1, m) = copy(ar0, n0, ar1, n1, m)
Proof step 27:
n < n1, n < max(# ar1, n1 + m0), # ar1 < n1 + m0 ⊦ copy(ar0, n0, resize(ar1, n1 + m0), n1, m0)[n] = (# ar1 ≤ n ⊃ zero;ar1[n]), m0 = 0

Rule argument

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

Additional information

Used simplifier rules: splice-copy: ⊦ n1 + m ≤ # ar1 ∧ m ≠ 0 → splice(ar0, n0, ar1, n1, m) = copy(ar0, n0, ar1, n1, m);% ⊦ m < n ↔ ¬ n ≤ m; (ss)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → max(n, m) = n; (s)%⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]); (s)% ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: m0, n1, ar1, n0, ar0