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Proof step 132 from proof lim-exists (previous step)

n = n0 +1, # σσ = ⌜n⌝ , mon(σσ) ⊦ σσ = [], ∃ σ. (∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ)) ∧ (∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%inf-inseq-reduce: ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%succ: ⊦ m +1 = m + 1; from specification nat-basic2