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Proof step 7 from proof subarray-append-shift-right (previous step)

n0 ≤ # ar1 ⊦ # subarray(ar0 + ar1, 0, n0 + # ar0) = #(ar0 + subarray(ar1, 0, n0)) ∧ (∀ n. n < # subarray(ar0 + ar1, 0, n0 + # ar0) → subarray(ar0 + ar1, 0, n0 + # ar0)[n] = (ar0 + subarray(ar1, 0, n0))[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 9:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 10:
n < n0 + # ar0, n0 ≤ # ar1 ⊦ (ar0 + ar1)[n] = (ar0 + subarray(ar1, 0, n0))[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% ⊦ n + 0 = n; (s)% ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s)%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)% ⊦ # subarray(ar, m, n) = n; (s)