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Proof step 15 from proof from-append-add (previous step)

n ≤ # x0 ⊦ (x0 to n) + x from # x + n = [], #(x0 to n) < # x + n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
n ≤ # x0 ⊦ (x0 to n) from n = []

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;% ⊦ n + 0 = n; (s)% ⊦ # [] = 0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ x + [] = x; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m ≤ n → (n < m ↔ false); (s)