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Proof step 44 from proof subarray-eq-elim (previous step)

# ar = (n1 + m) + n, subarray(ar, n, n1 + m) = subarray(ar, n, 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

Proof step 45:
⊦ n0 + n1 ≤ # ar0 ∧ m0 + m1 ≤ n1 → subarray(subarray(ar0, n0, n1), m0, m1) = subarray(ar0, n0 + m0, m1)
Proof step 46:
# ar = (n1 + m) + n, subarray(ar, n, n1 + m) = subarray(ar, n, m) + subarray(ar, n + m, n1), n + n1 + m ≤ # ar, m + n1 ≤ n1 + m ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

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] → [n1, m, n1 + m, n, ar] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m ≤ m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ 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 → ¬ n < m; (s) Used terms: n1, m, n1 + m, n, ar