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Proof step 2 from proof subarray-up-same (previous step)

n < # ar ⊦ copy(ar[n, a], 0, mkarray(n), 0, n) = copy(ar, 0, mkarray(n), 0, n), # ar = 0

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:
n < # ar ⊦ # copy(ar[n, a], 0, mkarray(n), 0, n) = # copy(ar, 0, mkarray(n), 0, n) ∧ (∀ n0. n0 < # copy(ar[n, a], 0, mkarray(n), 0, n) → copy(ar[n, a], 0, mkarray(n), 0, n)[n0] = copy(ar, 0, mkarray(n), 0, n)[n0]), # ar = 0

Rule argument

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

Additional information

Used terms: copy(ar, 0, mkarray(n), 0, n), copy(ar[n, a], 0, mkarray(n), 0, n)