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Proof step 2 from proof sub-size (previous step)

s ⊆ s0, # s0 < # s ⊦ # s = 0, s = ∅

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s

Rule argument

spec: set-basic inst: name: size-exgr seq: # s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s subst: [s, s0] → [s0, s] flags: current, []

Additional information

Used simplifier rules: ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; Used terms: s0, s