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Proof step 4 from proof copy-is-subarray (previous step)

⊦ # copy(ar0, n0, ar1, 0, # ar1) = # copy(ar0, n0, mkarray(# ar1), 0, # ar1) ∧ (∀ n. n < # copy(ar0, n0, ar1, 0, # ar1) → copy(ar0, n0, ar1, 0, # ar1)[n] = copy(ar0, n0, mkarray(# ar1), 0, # ar1)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < # ar1 ⊦ ar0[n + n0] = copy(ar0, n0, mkarray(# ar1), 0, # ar1)[n], # ar1 = 0

Rule argument

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%at-copy: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]);%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%sub-base: ⊦ m - 0 = m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%Size-mkarray: ⊦ # mkarray(n) = n; from specification array