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Proof step 5 from proof subarray-append-shift-left (previous step)

# ar0 = n0 + n ⊦ subarray(ar0 + ar1, n, n0 + # ar1) = subarray(ar0, n, n0) + ar1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])
Proof step 7:
# ar0 = n0 + n ⊦ # subarray(ar0 + ar1, n, n0 + # ar1) = #(subarray(ar0, n, n0) + ar1) ∧ (∀ n1. n1 < # subarray(ar0 + ar1, n, n0 + # ar1) → subarray(ar0 + ar1, n, n0 + # ar1)[n1] = (subarray(ar0, n, n0) + ar1)[n1])

Rule argument

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

Additional information

Used terms: subarray(ar0, n, n0) + ar1, subarray(ar0 + ar1, n, n0 + # ar1)