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

⊦ x0 = nf to n ∧ nf0 = nf from n ↔ nf = (⌜x0⌝ + ⌜nf0⌝ ).f ∧ # 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⌝ )
Proof step 3:
¬ # ⌜nf⌝ < ⌜n⌝ → (⌜x0⌝ = ⌜nf⌝ to n ∧ ⌜nf0⌝ = ⌜nf⌝ from n ↔ ⌜nf⌝ = ⌜x0⌝ + ⌜nf0⌝ ∧ # ⌜x0⌝ = ⌜n⌝ ) ⊦ x0 = nf to n ∧ nf0 = nf from n ↔ nf = (⌜x0⌝ + ⌜nf0⌝ ).f ∧ # x0 = 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] → [⌜nf⌝ , n, ⌜x0⌝ , ⌜nf0⌝ ] flags: current, []

Additional information

Used terms: ⌜nf⌝ , n, ⌜x0⌝ , ⌜nf0⌝