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Proof step 46 from proof range-eq-splice-less (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 47:
⊦ # resize(ar, n) = n
Proof step 48:
# 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] ⊦ m1 = 0, n1 + m1 ≤ # splice(ar, n, ar0, n0, m0) ∧ (n3 < m1 → ar2[n2 + n3] = splice(ar, n, ar0, n0, m0)[n1 + n3])

Rule argument

Additional information

Used simplifier rules: resize-len: ⊦ # resize(ar, n) = n;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ # copy(ar, m, ar0, m0, n) = # ar0; (s)