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Proof step 12 from proof size-one (previous step)

# s = 1, ∀ a. ¬ a ∈ s ⊦ a ∈ s, ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: set-basic inst: name: alle seq: ∀ a. ¬ a ∈ s ⊦ s = ∅ subst: [s] → [s] flags: current, []

Additional information

Used simplifier rules: Size-empty: ⊦ # ∅ = 0; Used terms: s