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Proof step 3 from proof copy-shift-right (previous step)

⊦ copy(ar, m, ar0 + ar1, n0 + # ar0, m1) = ar0 + copy(ar, m, ar1, n0, m1)

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(ar, m, ar0 + ar1, n0 + # ar0, m1) = #(ar0 + copy(ar, m, ar1, n0, m1)) ∧ (∀ n. n < # copy(ar, m, ar0 + ar1, n0 + # ar0, m1) → copy(ar, m, ar0 + ar1, n0 + # ar0, m1)[n] = (ar0 + copy(ar, m, ar1, n0, m1))[n])

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [ar0 + copy(ar, m, ar1, n0, m1), copy(ar, m, ar0 + ar1, n0 + # ar0, m1)] flags: <all paths>

Additional information

Used terms: ar0 + copy(ar, m, ar1, n0, m1), copy(ar, m, ar0 + ar1, n0 + # ar0, m1)