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Proof step 4 from proof dropn-first (previous step)

⊦ (n ≤ # x ⊃ restn(n, x);[]) = (x = [] ⊃ [];(n ≤ (# x - 1) + 1 ⊃ restn(n - 1, x.tail);[])), n = 0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
n ≤ # x ⊦ n = 0, restn(n, x) = (x = [] ⊃ [];(n ≤ (# x - 1) + 1 ⊃ restn(n - 1, x.tail);[]))

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ # x = 0 ↔ x = []; (s)%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic