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Proof step 28 from proof monls-mon-implies-monls (previous step)

((lim(nσf) to n + 2) + nσ)[n] < lim(nσf)[n], ((lim(nσf) to n + 2) + nσ)[n] < lim(nσf)[n + 1], lim(nσf)[n + 1] ≤ lim(nσf)[n], mon(nσf), ⌜n⌝ + ⌜1⌝ < # lim(nσf), ∀ n. nσf(n) ⊑ lim(nσf) ⊦ lim(nσf)[n + 1] = 0, lim(nσf)[n] = 0, # lim(nσf) < ⌜n⌝ + ⌜2⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: stream inst: natstream name: get-append-ls seq: ⊦ ⌜m⌝ < # nσ → (nσ + nσ0)[m] = nσ[m] subst: [m, nσ0, nσ] → [n, nσ, lim(nσf) to n + 2] flags: rewrite

Additional information

Used simplifier rules: add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%size-to: ¬ # nσ < ⌜n⌝ ⊦ #(nσ to n) = ⌜n⌝ ; from specification stream[natstream]%ls-natinf-to-nat-1-2: ⊦ ⌜n⌝ < ⌜m⌝ + ⌜n0⌝ ↔ n < m + n0; from specification nat-inf%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%get-to: m < n, ¬ # nσ < ⌜n⌝ ⊦ (nσ to n)[m] = nσ[m]; from specification stream[natstream]%ref-1: ⊦ ¬ n < n; from specification nat-basic1%get-append-notls: ⊦ ¬ ⌜m⌝ < # nσ → (nσ + nσ0)[m] = nσ0[m - (# nσ).n]; from specification stream[natstream]%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic%prime-add-fin: n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; from specification nat-inf%sel: ⊦ ⌜m⌝ .n = m; from specification nat-inf-basic%append-assoc: ⊦ (nσ + nσ0) + nσ1 = nσ + nσ0 + nσ1; from specification stream[natstream]%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat Used terms: n, nσ, lim(nσf) to n + 2