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Proof step 14 from proof splice-resize-less (previous step)

n < n2 + n1, n2 ≤ m0 ⊦ (# splice(ar0, n0, ar1, n1, m0) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, m0)[n]) = (# splice(ar0, n0, ar1, n1, n2) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, n2)[n]), m0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ m ≠ 0 → # splice(ar0, n0, ar1, n1, m) = max(# ar1, n1 + m)
Proof step 16:
n2 = 0, n < n2 + n1, n2 ≤ m0 ⊦ (# splice(ar0, n0, ar1, n1, m0) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, m0)[n]) = (# splice(ar0, n0, ar1, n1, n2) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, n2)[n]), m0 = 0
Proof step 32:
n < n2 + n1, n2 ≤ m0 ⊦ (# splice(ar0, n0, ar1, n1, m0) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, m0)[n]) = (max(# ar1, n1 + n2) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, n2)[n]), m0 = 0, n2 = 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] → [n2, n1, ar1, n0, ar0] flags: <all paths>

Additional information

Used terms: n2, n1, ar1, n0, ar0