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Proof step 3 from proof emptystream-cancel (previous step)

⊦ # σ = ∞, σ + σ⊥ = σ⊥ ↔ σ = σ⊥ to (# σ).n

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
# σ = ⌜n⌝ ⊦ # σ = ∞, σ + σ⊥ = σ⊥ ↔ σ = σ⊥ to n

Rule argument

spec: nat-inf name: elim seq: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ) subst: [n∞, m] → [# σ, m]

Additional information

Used simplifier rules: ⊦ n∞ ≠ ∞ → (n = n∞.n ↔ n∞ = ⌜n⌝ ); (elim) Used terms: # σ, m