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Proof step 2 from proof lastn-length-01 (previous step)

n ≤ # x ⊦ # lastn(n, x) = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: lastn-restn-01: n ≤ # x ⊦ lastn(n, x) = restn(# x - n, x);%rest-len: n ≤ # x ⊦ # restn(n, x) = # x - n;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0; (s)% ⊦ ¬ n < n; (s)% ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m ≤ n ↔ ¬ n < m; (ss)