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Proof step 3 from proof append-at (previous step)

n < # ar0 + # ar1 ⊦ (ar0 + ar1)[n] = (n < # ar0 ⊃ ar0[n];ar1[n - # ar0])

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: append-at1: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%append-at2: ⊦ # ar0 ≤ n ∧ n < # ar0 + # ar1 → (ar0 + ar1)[n] = ar1[n - # ar0];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m < n ↔ ¬ n ≤ m; (ss)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic