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Proof step 6 from proof infinite-append-lem (previous step)

∀ 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. ¬ n0 < m → 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 all left

And obtained the following new goals

Proof step 7:
∃ 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. ¬ n0 < m → 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

pos: left 1 terms: lf abort: F discard: T computed: T

Additional information

Used terms: lf