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Proof step 6 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), n0 < n ⊦ copy(ar0, 0, mkarray(n), 0, n)[n0] = copy(ar1, 0, mkarray(n), 0, n)[n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 8:
# copy(ar0, 0, mkarray(n + m), 0, n + m) = # copy(ar1, 0, mkarray(n + m), 0, n + m), n0 < n, ∀ n0. n0 < # copy(ar0, 0, mkarray(n + m), 0, n + m) → copy(ar0, 0, mkarray(n + m), 0, n + m)[n0] = copy(ar1, 0, mkarray(n + m), 0, n + m)[n0] ⊦ 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 + m), 0, n + m), copy(ar0, 0, mkarray(n + m), 0, n + m)] flags: <all paths>

Additional information

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