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Proof step 31 from proof infinite-append-butlast (previous step)

(σ 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, ⌜(n0 = 0 ⊃ 0;ii(n0) - 1) + # lf(n0) - 1⌝ < # σ, ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).first, ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ 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)⌝ , ∀ 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)) ⊦ m ≤ n1 → (m = 0 ⊃ 0;ii(m) - 1) ≤ (n1 = 0 ⊃ 0;ii(n1) - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
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), (σ 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⌝ < # σ, m ≤ n1, (n1 = 0 ⊃ 0;ii(n1) - 1) < ii(m) - 1, (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)), ∀ n. lf(n).last = lf(n + 1).first ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ 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)⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ n. σ[ii(n + 1) - 1] = lf(n + 1).first ⊦ 0 = ii(m) - 1, m = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat