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Proof step 56 from proof range-copy-inner (previous step)

n4 < n0 + n3 + m0, n1 + n4 < # ar0, ((n3 + n1) + m0) + n0 ≤ # ar0, (n3 + n2 + m0) + n0 ≤ # ar2, (n3 + n1) + m0 ≤ # ar0, (n3 + n2) + m0 ≤ # ar2, n + m0 ≤ # ar, n2 + n3 ≤ # ar2, n1 + n3 ≤ # ar0, n3 + m0 ≤ n4, ∀ n. n < n0 → ar2[(n3 + n2 + m0) + n] = ar0[((n3 + n1) + m0) + n], ∀ n4. n4 < m0 → ar2[(n3 + n2) + n4] = ar[n + n4], ∀ n. n < n3 → ar2[n2 + n] = ar0[n1 + n] ⊦ ar2[n2 + n4] = ar0[n1 + n4], n0 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 57:
⊦ m ≤ n ↔ (∃ n0. m + n0 = n)
Proof step 58:
n1 + n4 < # ar0, n4 < n0 + n3 + m0, n1 + n3 ≤ # ar0, n2 + n3 ≤ # ar2, n + m0 ≤ # ar, (n3 + n2) + m0 ≤ # ar2, (n3 + n1) + m0 ≤ # ar0, (n3 + n2 + m0) + n0 ≤ # ar2, ((n3 + n1) + m0) + n0 ≤ # ar0, ∀ n. n < n3 → ar2[n2 + n] = ar0[n1 + n], ∀ n4. n4 < m0 → ar2[(n3 + n2) + n4] = ar[n + n4], ∀ n. n < n0 → ar2[(n3 + n2 + m0) + n] = ar0[((n3 + n1) + m0) + n], ∃ n0. (n3 + m0) + n0 = n4 ⊦ ar2[n2 + n4] = ar0[n1 + n4], n0 = 0

Rule argument

spec: nat name: le-addex seq: ⊦ m ≤ n ↔ (∃ n0. m + n0 = n) subst: [n, m] → [n4, n3 + m0] flags: <all paths>

Additional information

Used terms: n4, n3 + m0