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Proof step 9 from proof copy-coalesce (previous step)

n3 < # ar1, n3 < m0, n2 < # ar2, m0 ≤ # ar1, m0 + n ≤ # ar1, n3 + n2 < # ar2 ⊦ m0 = 0, # ar2 = 0, copy(ar0, n0, ar1, 0, # ar1)[n3 + n] = ar0[n3 + n0 + n], n2 + m0 = 0, # ar1 = 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] → [n3 + n, # ar1, 0, ar1, n0, ar0] flags: rewrite, current

Additional information

Used simplifier rules: zero-01: ⊦ ¬ m < 0; from specification nat-basic1%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%sub-base: ⊦ m - 0 = m; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat Used terms: n3 + n, # ar1, 0, ar1, n0, ar0