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Proof step 2 from proof subarray-at (previous step)

n0 < n ⊦ copy(ar, m, mkarray(n), 0, n)[n0] = ar[m + n0], n = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0])

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, ar1, ar0, n, n0, m] → [n0, mkarray(n), ar, m, 0, n] flags: current, []

Additional information

Used simplifier rules: Size-mkarray: ⊦ # mkarray(n) = n; from specification array%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 Used terms: n0, mkarray(n), ar, m, 0, n