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Proof step 45 from proof infinite-append-butlast (previous step)

# x = ii(n), ii(0) = 0, ii(n0 + 1) = ((n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1) + 1, ii(n0 + 1) = ii(n0) + # (n0 = 0 ⊃ lf(0);lf(n0).rest), (x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).rest) = (n0 = 0 ⊃ lf(0);lf(n0).rest), (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 < ii(n), ii(n0 + 1) ≤ ii(n), ∀ n1. ii(n1 + 1) = ii(n1) + # (n1 = 0 ⊃ lf(0);lf(n1).rest) ∧ ¬ ii(n) < ii(n1 + 1) ∧ (x from ii(n1)) to # (n1 = 0 ⊃ lf(0);lf(n1).rest) = (n1 = 0 ⊃ lf(0);lf(n1).rest), ∀ n1. x[ii(n1 + 1) - 1] = lf(n1 + 1).first, ∀ m. n ≤ m → (m = 0 ⊃ lf(0);lf(m).rest) = [] ∧ ii(m) = ii(n), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. n ≠ 0 → ii(n) ≠ 0 ⊦ ii(n) = 0, ∃ n1. 1 + (n1 = 0 ⊃ 0;ii(n1) - 1) = ii(n) ∧ (∀ m. n1 ≤ m → lf(m).rest = [] ∧ (m = 0 ⊃ 0;ii(m) - 1) = (n1 = 0 ⊃ 0;ii(n1) - 1))

User applied rule exists right

And obtained the following new goals

Rule argument

pos: right 2 terms: n abort: F discard: T computed: F

Additional information

Used terms: n