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Proof step 24 from proof diag-theorem-butlast (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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