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Proof step 11 from proof split-prefix-append-to (previous step)

# σ0 = ∞, # σ1 = ∞, ⌜n⌝ + ⌜n0⌝ < # σ0, ∀ n1. ((σ0 to n) + σ1)[n1] = σ0[n1] ⊦ σ1[n0] = σ0[n + n0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# σ0 = ∞, # σ1 = ∞, ∀ n1. ((σ0 to n) + σ1)[n1] = σ0[n1] ⊦ σ1[n0] = σ0[n + n0]

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% ⊦ ⌜m⌝ ≠ ∞; (s)% ⊦ ⌜m⌝ + n∞ = ∞ ↔ n∞ = ∞; (s)% ⊦ n∞ < ∞ ↔ n∞ ≠ ∞; (s)