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Proof step 5 from proof split-prefix-append-to (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# σ0 = ∞, (σ0 to n) + σ1 = σ0 ⊦ # σ1 = ∞ ∧ (⌜n0⌝ < # σ1 → σ1[n0] = (σ0 from n)[n0])

Rule argument

Additional information

Used simplifier rules: ⊦ # σ0 = ∞ → #(σ0 from n) = ∞; (s)