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Proof step 10 from proof subarray-elim (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: subarray-append-consec0 seq: ⊦ n0 + n1 ≤ # ar → subarray(ar, 0, n0) + subarray(ar, n0, n1) = subarray(ar, 0, n0 + n1) subst: [n1, n0, ar] → [n0, n + m, ar] flags: <all paths>

Additional information

Used simplifier rules: append-prefix-left: ⊦ ar0 + ar1 = ar0 ↔ # ar1 = 0;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%# ar = n ⊦ subarray(ar, 0, n) = ar; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)% ⊦ # subarray(ar, m, n) = n; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → ¬ n < m; (s) Used terms: n0, n + m, ar