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Proof step 13 from proof diag-theorem-butlast (previous step)

∃ σ, ii. ii(0) = 0 ∧ (∀ 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)) ∧ (finp(σ) ⊃ ∃ n. # 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(0) = a ', ∀ n. lf(n + 1) = nextl(n, lf(n).last) ∧ lf(n) ≠ [], ∀ n, a. nextl(n, a) ≠ [] ∧ nextl(n, a).first = a ⊦ ∃ σ, ii. ii(0) = 0 ∧ σ[0] = a ∧ σ ≠ [] ∧ (∀ n. ii(n + 1) = ii(n) + # nextl(n, σ[ii(n)]) - 1 ∧ ⌜ii(n + 1)⌝ < # σ ∧ (σ from ii(n)) to # nextl(n, σ[ii(n)]) = ⌜nextl(n, σ[ii(n)])⌝ ) ∧ (finp(σ) ⊃ ∃ n. # σ = ⌜ii(n)⌝ + ⌜1⌝ ∧ (∀ m. n ≤ m → nextl(m, σ[ii(m)]) = σ[ii(m)] ' ∧ ii(n) = ii(m));∀ 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: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf