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Proof step 8 from proof subarray-eq-elim (previous step)

# ar = n0 + n, n + m ≤ # ar, n ≤ # ar, n < # ar → ar = subarray(ar, 0, n) + subarray(ar, n, n0) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# ar = n0 + n, m ≤ n0, n0 ≠ 0 → ar = subarray(ar, 0, n) + subarray(ar, n, n0) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m ≤ m + n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)