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Proof step 3 from proof size-exgr (previous step)

⊦ # s0 = 0, s0 = ∅, ∃ a. a ∈ s0

User applied rule constructor cut

And obtained the following new goals

Proof step 4:
s0 = s ++ a0 ⊦ # s0 = 0, s0 = ∅, s0 = s, ∃ a. a ∈ s0

Rule argument

var: s0

Additional information

Used simplifier rules: gen: ⊦ ;