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Proof step 12 from proof mon-lem (previous step)

σσ[m] + σ = σσ[m1 + m], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, ⌜m1⌝ + ⌜m⌝ < # σσ, ⌜m1 + m⌝ + ⌜1⌝ < # σσ → (∃ σ1. σσ[m1 + m] + σ1 = σσ[(m1 + m) + 1]), ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σσ[n + 1]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[(m1 + 1) + m]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
σσ[m] + σ = σσ[m1 + m], σσ[m1 + m] + σ0 = σσ[(m1 + m) + 1], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, ⌜m1⌝ + ⌜m⌝ < # σσ, ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σσ[n + 1]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[(m1 + 1) + m]

Rule argument

Additional information

Used simplifier rules: 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