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Proof step 25 from proof to-to-leq (previous step)

n0 < # x, m ≤ n0 + 1, m ≤ # x, m ≤ #(x to n0), ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ x to n0 to m = x to m, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
m < # x, n0 < # x, m ≤ n0, m ≤ n0 + 1, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ x to n0 to m = x to m, x = []

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;% ⊦ m < n → m ≤ n; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → ¬ n < m; (s)