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Proof step 3 from proof from-append-toinffrom (previous step)

n∞1 = ⌜n0⌝ ⊦ (σ to n) + ((σ from n) to∞ n∞1) = σ to∞ ⌜n⌝ + n∞1, # σ < ⌜n⌝ + n∞1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (σ to n) + ((σ from n) to n0) = σ to n0 + n, # σ < ⌜n⌝ + ⌜n0⌝

Rule argument

Additional information

Used simplifier rules: toinf-fin: ⊦ σ to∞ ⌜n⌝ = σ to n;%toinf-add: ⊦ σ to∞ ⌜n⌝ + ⌜m⌝ = σ to n + m;%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