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Proof step 3 from proof subarray-append-consec (previous step)

m0 + n0 + n1 ≤ # ar ⊦ subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1) = subarray(ar, m0, n0 + n1)

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:
m0 + n0 + n1 ≤ # ar ⊦ #(subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1)) = # subarray(ar, m0, n0 + n1) ∧ (∀ n. n < #(subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1)) → (subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1))[n] = subarray(ar, m0, n0 + n1)[n])

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [subarray(ar, m0, n0 + n1), subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1)] flags: <all paths>

Additional information

Used terms: subarray(ar, m0, n0 + n1), subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1)