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Proof step 24 from proof to-to-leq (previous step)

m ≤ #(x to n0), n0 < # x, n0 < # x, m ≤ n0 + 1, m ≤ # x, ∀ 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 25:
m < # x, m ≤ n0, m ≤ n0 + 1, n0 < # x, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ x = [], x to n0 to m = x to m

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;%lels-03: ⊦ m < n → m ≤ n; from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1