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Proof step 50 from proof lim-exists (previous step)

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

User applied rule cut formula

And closed the goal!

Rule argument

fma: n ≤ m0

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ⌜[]⌝ = []; (s)% ⊦ σ + [] = σ; (s)%append-assoc: ⊦ (σ + σ1) + σ2 = σ + σ1 + σ2; from specification stream% ⊦ # [] = 0; (s)% ⊦ n + 0 = n; (s)% ⊦ x to 0 = []; (s) Used formulas: n ≤ m0