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Proof step 22 from proof subarray-copy (previous step)

n1 + n3 < m1, m0 + m1 ≤ # ar1, (n1 + n3) + m0 < # ar1, n3 < m1, n1 < n2 ⊦ n3 + n2 = 0, ar0[n1 + n0] = copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n3, n2)[n1 + n3], # ar1 = 0, m1 = 0, n2 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: inst: name: at-copy seq: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]) subst: [m0, m, n0, ar1, n, ar0] → [n1 + n3, n2, n3, copy(ar1, m0, mkarray(m1), 0, m1), n0, ar0] flags: rewrite, current

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%eq-01: ⊦ m + n - m = n; from specification nat%Size-mkarray: ⊦ # mkarray(n) = n; from specification array%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat Used terms: n1 + n3, n2, n3, copy(ar1, m0, mkarray(m1), 0, m1), n0, ar0