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Proof step 16 from proof split-prefix-append-to (conclusion)

⊦ # σ0 = ∞ → ((σ0 to n) + σ1 = σ0 ↔ σ1 = σ0 from n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# σ0 = ∞, (σ0 to n) + σ1 = σ0 ⊦ σ1 = σ0 from n

Rule argument

Additional information

[Concretesimprules((Csimpseq( ⊦ (σ0 + σ3) + σ4 = σ0 + σ3 + σ4) Csimpseq(¬ # σ0 < ⌜n⌝ ⊦ (σ0 to n) + (σ0 from n) ≠ σ0 ↔ false) Csimpseq( ⊦ ¬ ∞ < n∞)))]