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Proof step 5 from proof copy-shift-right (previous step)

⊦ # copy(ar, m, ar0 + ar1, n0 + # ar0, m1) = #(ar0 + copy(ar, m, ar1, n0, m1)) ∧ (∀ n. n < # copy(ar, m, ar0 + ar1, n0 + # ar0, m1) → copy(ar, m, ar0 + ar1, n0 + # ar0, m1)[n] = (ar0 + copy(ar, m, ar1, n0, m1))[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 7:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 8:
n < # ar0 + # ar1 ⊦ copy(ar, m, ar0 + ar1, n0 + # ar0, m1)[n] = (ar0 + copy(ar, m, ar1, n0, m1))[n]

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # copy(ar, m, ar0, m0, n) = # ar0; (s)