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Proof step 53 from proof range-eq-splice-less (previous step)

# 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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 54:
# ar0 < n0 + m1, n3 < m1, n3 < m0, 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] ⊦ ar2[n2 + n3] = splice(ar, n, ar0, n0, m0)[n1 + n3]

Rule argument

Additional information

Used simplifier rules: fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ n ≤ n0 ∧ n1 ≤ m → n + n1 ≤ n0 + m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)