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Proof step 57 from proof lim-exists (previous step)

# σσ = ∞, σσ[n] = σσ[0] + ⌜nf to ii(n)⌝ , ii(0) = 0, mon(σσ), ∀ n. ii(n + 1) = ii(n) + # lf(n) ∧ (nf from ii(n)) to # lf(n) = lf(n), ∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ , ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1), ∀ n. σσ[n] = σσ[0] + ⌜nf to ii(n)⌝ ⊦ σσ = [], ∃ σ1. σσ[n] + σ1 = σσ[0] + ⌜nf⌝ , ∃ n. # σσ[n] = ∞

User applied rule insert equation

And obtained the following new goals

Proof step 58:
# σσ = ∞, ii(0) = 0, mon(σσ), ∀ n. σσ[n] = σσ[0] + ⌜nf to ii(n)⌝ , ∀ n. ∃ m. ii(m) ≤ n ∧ n < ii(m + 1), ∀ m, n. m ≤ n → ii(m) ≤ ii(n), ∀ n. σσ[n + 1] = σσ[n] + ⌜lf(n)⌝ , ∀ n. ii(n + 1) = ii(n) + # lf(n) ∧ (nf from ii(n)) to # lf(n) = lf(n) ⊦ σσ = [], ∃ n. # σσ[n] = ∞, ∃ σ1. (σσ[0] + ⌜nf to ii(n)⌝ ) + σ1 = σσ[0] + ⌜nf⌝

Rule argument

pos: left 2 flags: discard

Proof step 57 from proof lim-exists (previous step)

User applied rule insert equation

And obtained the following new goals

Rule argument

Additional information