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Proof step 7 from proof copy-append (previous step)

n < # ar2, n0 + m1 ≤ # ar1 ⊦ (n < m0 ∨ m0 + m1 ≤ n ⊃ ar2[n];(ar0 + ar1)[n + n0 + # ar0 - m0]) = (n < m0 ∨ m0 + m1 ≤ n ⊃ ar2[n];ar1[n + n0 - m0])

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-atp2: n < # ar1 ⊦ (ar0 + ar1)[# ar0 + n] = ar1[n];%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)