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Proof step 6 from proof prefixn-append-slice (previous step)

n ≤ n0 ⊦ (n ≤ # x ⊃ firstn(n, x);x) + (n0 ≤ # (n ≤ # x ⊃ restn(n, x);[]) + n ⊃ firstn(n0 - n, (n ≤ # x ⊃ restn(n, x);[]));(n ≤ # x ⊃ restn(n, x);[])) = (n0 ≤ # x ⊃ firstn(n0, x);x)

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: ⊦ # [] = 0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m - m = 0; (s)% ⊦ firstn(0, x) = []; (s)% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ m < n → (n ≤ m ↔ false); (s)