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Proof step 6 from proof subarray-split-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# ar = n0 + n ⊦ ar = subarray(ar, 0, n) + subarray(ar, n, n0), n0 = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → m ≤ n; (s)