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Proof step 57 from proof infinite-append-butlast (previous step)

(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 = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: list-data inst: name: elim-list-c seq: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y subst: [x, a, y] → [lf(m), a0, y0] flags:

Additional information

Used simplifier rules: x ≠ [] ⊦ a = x.first ∧ y = x.rest ↔ x = a + y; (elim) Used terms: lf(m), a0, y0