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Proof step 5 from proof emptystream-cancel (previous step)

# σ = ⌜n⌝ ⊦ σ + σ⊥ = σ⊥ ↔ σ = σ⊥ to n

User applied rule case distinction

And obtained the following new goals

Proof step 6:
# σ = ⌜n⌝ , σ + σ⊥ = σ⊥ ⊦ σ = σ⊥ to n
Proof step 24:
# σ = ⌜n⌝ , σ = σ⊥ to n ⊦ σ + σ⊥ = σ⊥

Rule argument

pos: right 1

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