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Proof step 11 from proof to-to-leq (previous step)

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) + x[n0] ' to m = x to m, x = [], m ≤ #(x to n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n0 < # x, ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ (x to n0) + x[n0] ' to n0 + 1 = x to n0 + 1, x = []

Rule argument

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m + 1 ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s)