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Proof step 1 from proof elim-fromto-fin (conclusion)

⊦ ¬ # x < n → (x0 = x to n ∧ x1 = x from n ↔ x = x0 + x1 ∧ # x0 = n)

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
⊦ ¬ # σ < ⌜n⌝ → (σ0 = σ to n ∧ σ1 = σ from n ↔ σ = σ0 + σ1 ∧ # σ0 = ⌜n⌝ )

Rule argument

spec: inst: name: elim-fromto seq: ⊦ ¬ # σ < ⌜n⌝ → (σ0 = σ to n ∧ σ1 = σ from n ↔ σ = σ0 + σ1 ∧ # σ0 = ⌜n⌝ ) subst: [σ, n, σ0, σ1] → [⌜x⌝ , n, ⌜x0⌝ , ⌜x1⌝ ] flags: current, []

Additional information

Used simplifier rules: length-to: ¬ # x < n ⊦ #(x to n) = n;%append-to-size: ⊦ x + y to # x = x;%append-from-size: ⊦ x + y from # x = y;%to-plus-from-fin: ⊦ n ≤ # x → (x to n) + (x from n) = x;%a: ⊦ (x + y) + z = x + y + z; from specification list%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat Used terms: ⌜x⌝ , n, ⌜x0⌝ , ⌜x1⌝