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Proof step 14 from proof from-append-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
n ≤ # x0 ⊦ (x0 to n) from n = []

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;%add-base: ⊦ n + 0 = n; from specification nat-basic2%size-01: ⊦ # [] = 0; from specification list-data%a: ⊦ (x + y) + z = x + y + z; from specification list%app: ⊦ x + [] = x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ # x = 0 ↔ x = []; from specification list-data%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat