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Proof step 42 from proof range-extend (previous step)

n0 + n2 < # ar0, n2 < m0 + m1, n0 + m0 + m1 ≤ # ar0, n0 + m0 ≤ # ar0, n1 + m0 ≤ # ar1, n + m1 ≤ # ar, (n1 + m0) + m1 ≤ # ar1, ∀ n. n < m0 → ar0[n0 + n] = ar1[n1 + n], ∀ n2. n2 < m1 → ar[n + n2] = ar1[(n1 + m0) + n2] ⊦ ar[n2 + n - m0] = ar1[n1 + n2], n2 < m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
n0 + n2 < # ar0, n2 < m0 + m1, m0 ≤ n2, n0 + m0 + m1 ≤ # ar0, n0 + m0 ≤ # ar0, n1 + m0 ≤ # ar1, n + m1 ≤ # ar, (n1 + m0) + m1 ≤ # ar1, ∀ n. n < m0 → ar0[n0 + n] = ar1[n1 + n], ∀ n2. n2 < m1 → ar[n + n2] = ar1[(n1 + m0) + n2] ⊦ ar[(n2 - m0) + n] = ar1[n1 + n2], m1 = 0

Rule argument

Additional information

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