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Proof step 158 from proof infinite-append-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat