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Proof step 29 from proof fill-append-shift (previous step)

n < # ar0 + # ar1, n < # ar0 ⊦ (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;(ar0 + ar1)[n]) = ar0[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]
Proof step 31:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 32:
n < # ar0 + # ar1, n < # ar0 ⊦ (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;ar0[n]) = ar0[n]

Rule argument

Additional information

Used simplifier rules: append-at1: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic