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Proof step 3 from proof subarray-append-consec0 (previous step)

n0 + n1 ≤ # ar ⊦ subarray(ar, 0, n0) + subarray(ar, n0, n1) = subarray(ar, 0, n0 + n1)

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: subarray-append-consec seq: ⊦ m0 + n0 + n1 ≤ # ar → subarray(ar, m0, n0) + subarray(ar, n0 + m0, n1) = subarray(ar, m0, n0 + n1) subst: [m0, n0, n1, ar] → [0, n0, n1, ar]

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: 0, n0, n1, ar