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Proof step 38 from proof range-extend (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
n2 < m0 + m1, n0 + n2 < # ar0, (n1 + m0) + m1 ≤ # ar1, n + m1 ≤ # ar, n1 + m0 ≤ # ar1, n0 + m0 ≤ # ar0, n0 + m0 + m1 ≤ # ar0, ∀ n2. n2 < m1 → ar[n + n2] = ar1[(n1 + m0) + n2], ∀ n. n < m0 → ar0[n0 + n] = ar1[n1 + n] ⊦ (n2 < m0 ⊃ ar0[n0 + n2];ar[n2 + n - m0]) = ar1[n1 + n2]

Rule argument

Additional information

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