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Proof step 43 from proof range-copy-lower (previous step)

(n3 + n0) + n5 < # ar0, n5 < n4 + n1, n0 ≤ # ar0, n3 + n0 ≤ # ar0, n0 + n1 + n3 ≤ # ar0, n + n1 + n3 ≤ # ar, n2 + n1 ≤ # ar2, (n + n3) + n1 ≤ # ar, (n1 + n2) + n4 ≤ # ar2, (n0 + n1 + n3) + n4 ≤ # ar0, n1 ≤ n5, ∀ n4. n4 < n1 → ar2[n2 + n4] = ar[(n + n3) + n4], ∀ n. n < n4 → ar2[(n1 + n2) + n] = ar0[(n0 + n1 + n3) + n] ⊦ ar2[n2 + n5] = ar0[(n3 + n0) + n5], n4 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 44:
⊦ m ≤ n ↔ (∃ n0. m + n0 = n)
Proof step 45:
n5 < n4 + n1, (n3 + n0) + n5 < # ar0, (n0 + n1 + n3) + n4 ≤ # ar0, (n1 + n2) + n4 ≤ # ar2, (n + n3) + n1 ≤ # ar, n2 + n1 ≤ # ar2, n + n1 + n3 ≤ # ar, n0 + n1 + n3 ≤ # ar0, n3 + n0 ≤ # ar0, n0 ≤ # ar0, ∀ n. n < n4 → ar2[(n1 + n2) + n] = ar0[(n0 + n1 + n3) + n], ∀ n4. n4 < n1 → ar2[n2 + n4] = ar[(n + n3) + n4], ∃ n0. n1 + n0 = n5 ⊦ ar2[n2 + n5] = ar0[(n3 + n0) + n5], n4 = 0

Rule argument

spec: nat name: le-addex seq: ⊦ m ≤ n ↔ (∃ n0. m + n0 = n) subst: [n, m] → [n5, n1] flags: <all paths>

Additional information

Used terms: n5, n1