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Proof step 18 from proof splice-resize-less (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ m ≠ 0 → # splice(ar0, n0, ar1, n1, m) = max(# ar1, n1 + m)
Proof step 20:
⊦ splice(ar0, n0, ar1, n1, 0) = ar1
Proof step 21:
n < n1 ⊦ (max(# ar1, n1 + m0) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, m0)[n]) = (# ar1 ≤ n ⊃ zero;ar1[n]), 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, n1, ar1, n0, ar0] flags: <all paths>

Additional information

Used simplifier rules: splice-0: ⊦ splice(ar0, n0, ar1, n1, 0) = ar1; Used terms: m0, n1, ar1, n0, ar0