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Proof step 111 from proof infinite-append-butlast (previous step)

ii(n1 + 1) = m0 + 1 ∧ ii(n1 + 1) ≠ 0, ii(0) = 0, σ[m0] = lf(n1 + 1).first, ii(2 + n1) = ii(n1 + 1) + # lf(n1 + 1).rest, (σ from ii(n1 + 1)) to # lf(n1 + 1).rest = ⌜lf(n1 + 1).rest⌝ , ∀ n. n ≠ 0 → ii(n) ≠ 0, ∀ 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)) ⊦ σ[ii(2 + n1) - 1] = lf(2 + n1).first, # σ < ⌜ii(2 + n1)⌝

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 112:
(σ from m0 + 1) to # lf(n1 + 1).rest = ⌜lf(n1 + 1).rest⌝ , ii(2 + n1) = (m0 + 1) + # lf(n1 + 1).rest, σ[m0] = lf(n1 + 1).first, ii(0) = 0, ii(n1 + 1) = m0 + 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. n ≠ 0 → ii(n) ≠ 0 ⊦ σ[ii(2 + n1) - 1] = lf(2 + n1).first, # σ < ⌜ii(2 + n1)⌝

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2