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Proof step 43 from proof range-extend (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 44:
n2 = n3 + m0, n2 < m0 + m1, n0 + n2 < # ar0, (n1 + m0) + m1 ≤ # ar1, n + m1 ≤ # ar, n1 + m0 ≤ # ar1, n0 + m0 ≤ # ar0, n0 + m0 + m1 ≤ # ar0, m0 ≤ n2, ∀ n2. n2 < m1 → ar[n + n2] = ar1[(n1 + m0) + n2], ∀ n. n < m0 → ar0[n0 + n] = ar1[n1 + n] ⊦ ar[n3 + n] = ar1[n1 + n2], m1 = 0

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n2, m0, n3]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n2, m0, n3