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Proof step 2 from proof subarray-copy-lower (previous step)

n1 + n2 ≤ m0, m0 + m1 ≤ # ar ⊦ copy(copy(ar0, n0, ar, n1, n2), m0, mkarray(m1), 0, m1) = copy(ar, m0, mkarray(m1), 0, m1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 4:
n1 + n2 ≤ m0, m0 + m1 ≤ # ar ⊦ # copy(copy(ar0, n0, ar, n1, n2), m0, mkarray(m1), 0, m1) = # copy(ar, m0, mkarray(m1), 0, m1) ∧ (∀ n. n < # copy(copy(ar0, n0, ar, n1, n2), m0, mkarray(m1), 0, m1) → copy(copy(ar0, n0, ar, n1, n2), m0, mkarray(m1), 0, m1)[n] = copy(ar, m0, mkarray(m1), 0, m1)[n])

Rule argument

spec: array inst: name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [copy(ar, m0, mkarray(m1), 0, m1), copy(copy(ar0, n0, ar, n1, n2), m0, mkarray(m1), 0, m1)] flags: rewrite

Additional information

Used terms: copy(ar, m0, mkarray(m1), 0, m1), copy(copy(ar0, n0, ar, n1, n2), m0, mkarray(m1), 0, m1)