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Proof step 7 from proof expand-cons (previous step)

⊦ ⌜nf⌝ = [], ∃ a, σ0. ⌜nf⌝ = ⌜a '⌝ + σ0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ ∃ a, σ0. ⌜nf⌝ = ⌜a '⌝ + σ0

Rule argument

Additional information

Used simplifier rules: fun-nonempty: ⊦ [] ≠ ⌜nf⌝ ;