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

# s = 1 ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ s = ∅ ∨ (∃ a. a ∈ s)
Proof step 4:
s = ∅ ∨ (∃ a. a ∈ s), # s = 1 ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

Rule argument

spec: set-basic inst: name: sel seq: ⊦ s = ∅ ∨ (∃ a. a ∈ s) subst: [s] → [s] flags: current, []

Additional information

Used terms: s