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

⊦ copy(ar0, n0, ar1, 0, # ar1) = copy(ar0, n0, mkarray(# ar1), 0, # ar1)

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:
⊦ # copy(ar0, n0, ar1, 0, # ar1) = # copy(ar0, n0, mkarray(# ar1), 0, # ar1) ∧ (∀ n. n < # copy(ar0, n0, ar1, 0, # ar1) → copy(ar0, n0, ar1, 0, # ar1)[n] = copy(ar0, n0, mkarray(# ar1), 0, # ar1)[n])

Rule argument

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

Additional information

Used terms: copy(ar0, n0, mkarray(# ar1), 0, # ar1), copy(ar0, n0, ar1, 0, # ar1)