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Proof step 35 from proof subarray-eq-elim (previous step)

# ar = (n1 + m) + n, ar = subarray(ar, 0, n) + subarray(ar, n, n1 + m), subarray(ar, n, n1 + m) = subarray(subarray(ar, n, n1 + m), 0, m) + subarray(subarray(ar, n, n1 + m), m, n1) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: array-copy name: subarray-subarray-elim seq: ⊦ n0 + n1 ≤ # ar0 ∧ m0 + m1 ≤ n1 → subarray(subarray(ar0, n0, n1), m0, m1) = subarray(ar0, n0 + m0, m1) subst: [m1, m0, n1, n0, ar0] → [m, 0, n1 + m, n, 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;%append-zero-left: ⊦ mkarray(0) + ar = ar;%subarray-elim-right: n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m);% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ # subarray(ar, m, n) = n; (s)% ⊦ m = m + n ↔ n = 0; (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%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ ¬ n < n; (s)%# ar = n ⊦ subarray(ar, 0, n) = ar; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)% ⊦ subarray(ar, n, 0) = mkarray(0); (s)% ⊦ # mkarray(n) = n; (s)% ⊦ m - 0 = m; (s)% ⊦ # ar = 0 → (mkarray(0) = ar ↔ true); (ss)% ⊦ m < n → ¬ n < m; (s) Used terms: m, 0, n1 + m, n, ar