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Proof step 51 from proof range-eq-splice-less (previous step)

# ar0 < n0 + m1, n2 + m1 ≤ # ar2, n + m0 ≤ # ar, m1 ≤ m0, n1 ≤ n0, ∀ n3. n3 < m1 → ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m1), n0, m1)[n1 + n3] ⊦ m0 = 0, m1 = 0, n1 + m1 ≤ max(# ar0, n0 + m0) ∧ (n3 < m1 → ar2[n2 + n3] = splice(ar, n, ar0, n0, m0)[n1 + n3])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 52:
⊦ max(n, m) = (n < m ⊃ m;n)
Proof step 53:
# ar0 < n0 + m1, n1 ≤ n0, m1 ≤ m0, n + m0 ≤ # ar, n2 + m1 ≤ # ar2, ∀ n3. n3 < m1 → ar2[n2 + n3] = copy(ar, n, resize(ar0, n0 + m1), n0, m1)[n1 + n3] ⊦ m0 = 0, m1 = 0, n1 + m1 ≤ (# ar0 < n0 + m0 ⊃ n0 + m0;# ar0) ∧ (n3 < m1 → ar2[n2 + n3] = splice(ar, n, ar0, n0, m0)[n1 + n3])

Rule argument

spec: min-max name: max seq: ⊦ max(n, m) = (n < m ⊃ m;n) subst: [m, n] → [n0 + m0, # ar0] flags: <all paths>

Additional information

Used terms: n0 + m0, # ar0