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Proof step 17 from proof diag-theorem-butlast (previous step)

ii(0) = 0, ii(0 + 1) = ii(0) + # lf(0 + 1) - 1, lf(0) = a ', (σ from ii(0)) to # lf(0 + 1) = ⌜lf(0 + 1)⌝ , ⌜ii(0 + 1)⌝ < # σ, (finp(σ) ⊃ # σ = ⌜ii(n)⌝ + ⌜1⌝ ∧ (∀ m. n ≤ m → lf(m + 1).tail = [] ∧ ii(m) = ii(n));∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1)), ∀ n, a. nextl(n, a) ≠ [] ∧ nextl(n, a).head = a, ∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [], ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ii(n + 1) = ii(n) + # lf(n + 1) - 1 ∧ ⌜ii(n + 1)⌝ < # σ ∧ (σ from ii(n)) to # lf(n + 1) = ⌜lf(n + 1)⌝ ⊦ σ[0] = a ∧ σ ≠ [] ∧ ii(n0 + 1) = ii(n0) + # nextl(n0, σ[ii(n0)]) - 1 ∧ ⌜ii(n0 + 1)⌝ < # σ ∧ (σ from ii(n0)) to # nextl(n0, σ[ii(n0)]) = ⌜nextl(n0, σ[ii(n0)])⌝ ∧ (finp(σ) ⊃ ∃ n. # σ = ⌜ii(n)⌝ + ⌜1⌝ ∧ (∀ m. n ≤ m → nextl(m, σ[ii(m)]) = σ[ii(m)] ' ∧ ii(n) = ii(m));∃ m. ii(m) ≤ n1 ∧ n1 < ii(m + 1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < 0; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ # [] = ⌜0⌝ ; (s)% ⊦ σ from 0 = σ; (s)% ⊦ n + 0 = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic