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Proof step 40 from proof lim-exists (previous step)

# σσ[n] = ⌜n0⌝ , ii(0) = 0, # σσ = ∞, # x = ii(n), n ≤ m, 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[m] = σσ[n], σσ = [], ∃ n. # σσ[n] = ∞, ∃ σ. (∀ n. ∃ σ1. σσ[n] + σ1 = σ) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n, σ1. (σ to m) + σ1 = σσ[n]))

User applied rule weakening

And obtained the following new goals

Proof step 41:
# σσ[n] = ⌜n0⌝ , ii(0) = 0, # σσ = ∞, # x = ii(n), n ≤ m, 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[m] = σσ[n], σσ = [], ∃ n. # σσ[n] = ∞

Rule argument

poslist: right 4

Proof step 40 from proof lim-exists (previous step)

User applied rule weakening

And obtained the following new goals

Rule argument

Additional information