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Proof step 1 from proof size-exgr (conclusion)

# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s

User applied rule structural induction

And obtained the following new goals

Proof step 2:
# ∅ < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ ∅
Proof step 6:
∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1), #(s1 ++ a) < # s0 ⊦ ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1 ++ a

Rule argument

indvar: s

Additional information

Used simplifier rules: gen: ⊦ ;