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Proof step 21 from proof to-to-leq (previous step)

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 = x to n0, x = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ n ≤ # x → x + y to n = x to n
Proof step 23:
n0 < # x, n0 ≤ #(x to n0), ∀ m, x. m ≤ n0 ∧ n0 ≤ # x ∧ m ≤ # x → x to n0 to m = x to m ⊦ x to n0 to n0 = x to n0, x = []

Rule argument

spec: <toplevel> name: append-to-le seq: ⊦ n ≤ # x → x + y to n = x to n subst: [n, y, x] → [n0, x[n0] ', x to n0] flags: <all paths>

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;% ⊦ m < n → ¬ n < m; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ n < n; (s) Used terms: n0, x[n0] ', x to n0