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Proof step 5 from proof subarray-eq-elim (previous step)

n + m ≤ # ar ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ n < # ar → ar = subarray(ar, 0, n) + subarray(ar, n, # ar - n)
Proof step 7:
n + m ≤ # ar, n < # ar → ar = subarray(ar, 0, n) + subarray(ar, n, # ar - n) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

Rule argument

spec: <toplevel> name: subarray-split-at seq: ⊦ n < # ar → ar = subarray(ar, 0, n) + subarray(ar, n, # ar - n) subst: [n, ar] → [n, ar]

Additional information

Used terms: n, ar