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Proof step 61 from proof infinite-append-butlast (previous step)

ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).rest), ii(n0 + 1) = ((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1, ii(0) = 0, (nf from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).rest) = (n0 = 0 ⊃ lf(0);lf(n0).rest), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1), ∀ n. nf(ii(n + 1) - 1) = lf(n + 1).first, ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).rest) ∧ (nf from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).rest) = (n = 0 ⊃ lf(0);lf(n).rest) ⊦ ∃ m0. (m0 = 0 ⊃ 0;ii(m0) - 1) ≤ n2 ∧ n2 < ii(m0 + 1) - 1

User applied rule all left

And obtained the following new goals

Proof step 62:
ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).rest), ii(n0 + 1) = ((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1, ii(0) = 0, (nf from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).rest) = (n0 = 0 ⊃ lf(0);lf(n0).rest), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], ∃ m. ii(m) ≤ n2 + 1 ∧ n2 + 1 < ii(m + 1), ∀ n. nf(ii(n + 1) - 1) = lf(n + 1).first, ∀ n. ii(n + 1) = ii(n) + # (n = 0 ⊃ lf(0);lf(n).rest) ∧ (nf from ii(n)) to # (n = 0 ⊃ lf(0);lf(n).rest) = (n = 0 ⊃ lf(0);lf(n).rest) ⊦ ∃ m0. (m0 = 0 ⊃ 0;ii(m0) - 1) ≤ n2 ∧ n2 < ii(m0 + 1) - 1

Rule argument

pos: left 8 terms: n2 + 1 abort: F discard: T computed: F

Additional information

Used terms: n2 + 1