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

⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b)

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ # s ≠ 0 → (∃ a. a ∈ s)
Proof step 4:
# s ≠ 0 → (∃ a. a ∈ s) ⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b)

Rule argument

spec: <toplevel> name: selsize seq: ⊦ # s ≠ 0 → (∃ a. a ∈ s) subst: [s] → [s]

Additional information

Used terms: s