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Proof step 76 from proof lim-unique (previous step)

⌜m⌝ = # σ, ⌜m0⌝ = (⌜m⌝ + # σ1) + # σ2, ⌜n⌝ < # σσ, mon(σσ), m0 < m, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ m1. ¬ m0 < m1 → (∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ1. (σσ[n] + σ2 to m1) + σ1 = σσ[n0])), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])) ⊦ σσ = [], m = m0, m = 0

User applied rule structural induction

And obtained the following new goals

Proof step 77:
⌜m1⌝ = # σ2, ⌜m⌝ = # σ, ⌜m0⌝ = (⌜m⌝ + # σ1) + # σ2, ⌜n⌝ < # σσ, mon(σσ), m0 < m, ∀ n. ⌜n⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σ), ∀ m1. ¬ m0 < m1 → (∃ n0. ⌜n0⌝ < # σσ ∧ (∃ σ1. (σσ[n] + σ2 to m1) + σ1 = σσ[n0])), ∀ m. ¬ # σ < ⌜m⌝ → (∃ n. ⌜n⌝ < # σσ ∧ (∃ σ1. (σ to m) + σ1 = σσ[n])) ⊦ σσ = [], m = m0, m = 0

Rule argument

term: # σ2

Additional information

Used simplifier rules: natinf freely generated by ∞, ⌜ ⌝;;%add-inf2: ⊦ ⌜m⌝ + ∞ = ∞; from specification nat-inf%add-all-inf: ⊦ n∞ + ∞ = ∞; from specification nat-inf%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%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic