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Proof step 5 from proof dropn-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n ≤ # x ⊦ n = 0, restn(n, x) = (x = [] ⊃ [];restn(n - 1, x.tail))

Rule argument

Additional information

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