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Proof step 36 from proof splice-resize-less (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 37:
⊦ # ar1 < n1 + m ∧ m ≠ 0 → splice(ar0, n0, ar1, n1, m) = copy(ar0, n0, resize(ar1, n1 + m), n1, m)
Proof step 38:
n < n2 + n1, n2 ≤ m0 ⊦ (max(# ar1, n1 + m0) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, m0)[n]) = splice(ar0, n0, ar1, n1, n2)[n], m0 = 0, n2 = 0, # ar1 < n1 + n2 ∧ n2 ≠ 0
Proof step 50:
n < n2 + n1, # ar1 < n1 + n2, n2 ≤ m0 ⊦ (max(# ar1, n1 + m0) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, m0)[n]) = copy(ar0, n0, resize(ar1, n1 + n2), n1, n2)[n], m0 = 0, n2 = 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] → [n2, n1, ar1, n0, ar0] flags: <all paths>

Additional information

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