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Proof step 3 from proof subarray-longer-eq-eq (previous step)

copy(ar0, 0, mkarray(n + m), 0, n + m) = copy(ar1, 0, mkarray(n + m), 0, n + m) ⊦ copy(ar0, 0, mkarray(n), 0, n) = copy(ar1, 0, mkarray(n), 0, n)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 5:
copy(ar0, 0, mkarray(n + m), 0, n + m) = copy(ar1, 0, mkarray(n + m), 0, n + m) ⊦ # copy(ar0, 0, mkarray(n), 0, n) = # copy(ar1, 0, mkarray(n), 0, n) ∧ (∀ n0. n0 < # copy(ar0, 0, mkarray(n), 0, n) → copy(ar0, 0, mkarray(n), 0, n)[n0] = copy(ar1, 0, mkarray(n), 0, n)[n0])

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [copy(ar1, 0, mkarray(n), 0, n), copy(ar0, 0, mkarray(n), 0, n)] flags: <all paths>

Additional information

Used terms: copy(ar1, 0, mkarray(n), 0, n), copy(ar0, 0, mkarray(n), 0, n)