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Proof step 56 from proof infinite-append-butlast (previous step)

m ≠ 0, lf(m).rest = [], ii(m) = m0 + 1, lf(m) = a ' + y, lf(m).last = lf(m + 1).first, (x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).rest) = (n0 = 0 ⊃ lf(0);lf(n0).rest), 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, # x = m0 + 1, ii(n) = m0 + 1, a ∈ lf(m), ii(n0 + 1) ≤ m0 + 1, n ≤ m, (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 ≤ m0, ∀ n1. x[ii(n1 + 1) - 1] = lf(n1 + 1).first, ∀ n1. ii(n1 + 1) = ii(n1) + # (n1 = 0 ⊃ lf(0);lf(n1).rest) ∧ ¬ m0 + 1 < ii(n1 + 1) ∧ (x from ii(n1)) to # (n1 = 0 ⊃ lf(0);lf(n1).rest) = (n1 = 0 ⊃ lf(0);lf(n1).rest), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n) ⊦ lf(m) = [], y = [] ∧ (m = 0 ⊃ 0;m0) = m0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
(x from ii(n0)) to # (n0 = 0 ⊃ lf(0);lf(n0).rest) = (n0 = 0 ⊃ lf(0);lf(n0).rest), 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, # x = m0 + 1, ii(n) = m0 + 1, lf(m).rest = [], ii(m) = m0 + 1, lf(m) = a ' + y, lf(m).first = lf(m + 1).first, a ∈ lf(m), ii(n0 + 1) ≤ m0 + 1, n ≤ m, (n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1 ≤ m0, ∀ n1. x[ii(n1 + 1) - 1] = lf(n1 + 1).first, ∀ n1. ii(n1 + 1) = ii(n1) + # (n1 = 0 ⊃ lf(0);lf(n1).rest) ∧ ¬ m0 + 1 < ii(n1 + 1) ∧ (x from ii(n1)) to # (n1 = 0 ⊃ lf(0);lf(n1).rest) = (n1 = 0 ⊃ lf(0);lf(n1).rest), ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ m, n. m ≤ n → ii(m) ≤ ii(n) ⊦ lf(m) = [], m = 0, y = []

Rule argument

Additional information

Used simplifier rules: l-01: ⊦ x ≠ [] ∧ x.rest = [] → x.last = x.first; from specification list-last