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Proof step 2 from proof infinite-append-lem (previous step)

⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))

User applied rule structural induction

And obtained the following new goals

Proof step 3:
⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < 0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))
Proof step 5:
∀ lf. ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n0 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m))) ⊦ ∃ nf, ii. ii(0) = 0 ∧ (∀ m. m < n0 +1 → ii(m + 1) = ii(m) + # lf(m) ∧ (nf from ii(m)) to # lf(m) = lf(m) ∧ (∀ m0. m0 ≤ m → ii(m0) ≤ ii(m)))

Rule argument

indvar: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;