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Proof step 42 from proof lim-exists (previous step)

∀ σσ, n, n0, ii, x, lf. # σσ[n] = ⌜n0⌝ ∧ ii(0) = 0 ∧ # σσ = ∞ ∧ # x = ii(n) ∧ n ≤ m0 ∧ mon(σσ) ∧ (∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n)) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ ) ∧ (∀ n0. ii(n0 + 1) = ii(n0) + # lf(n0) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # lf(n0) = lf(n0)) → σσ[m0] = σσ[n] ∨ σσ = [] ∨ (∃ n. # σσ[n] = ∞), # σσ[n] = ⌜n0⌝ , ii(0) = 0, # σσ = ∞, # x = ii(n), n ≤ m0 +1, mon(σσ), ∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ , ∀ n0. ii(n0 + 1) = ii(n0) + # lf(n0) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # lf(n0) = lf(n0) ⊦ σσ[m0 +1] = σσ[n], σσ = [], ∃ n. # σσ[n] = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
# σσ[n] = ⌜n0⌝ , ii(0) = 0, # σσ = ∞, # x = ii(n), mon(σσ), n ≤ m0 + 1, ∀ σσ, n, n0, ii, x, lf. # σσ[n] = ⌜n0⌝ ∧ ii(0) = 0 ∧ # σσ = ∞ ∧ # x = ii(n) ∧ n ≤ m0 ∧ mon(σσ) ∧ (∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n)) ∧ (∀ m, n. m ≤ n → ii(m) ≤ ii(n)) ∧ (∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ ) ∧ (∀ n0. ii(n0 + 1) = ii(n0) + # lf(n0) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # lf(n0) = lf(n0)) → σσ[m0] = σσ[n] ∨ σσ = [] ∨ (∃ n. # σσ[n] = ∞), ∀ m. n ≤ m → lf(m) = [] ∧ ii(m) = ii(n), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ , ∀ n0. ii(n0 + 1) = ii(n0) + # lf(n0) ∧ ¬ ii(n) < ii(n0 + 1) ∧ (x from ii(n0)) to # lf(n0) = lf(n0) ⊦ σσ = [], σσ[m0 + 1] = σσ[n], ∃ n. # σσ[n] = ∞

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2