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Proof step 15 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

Rule argument

spec: array inst: 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: rewrite

Additional information

[Concretesimprules(()), Extraterms((subarray(ar, m0, n0 + n1) subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1)))]