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Proof step 1 from proof size-three-02 (conclusion)

⊦ # s = 3 ↔ (∃ a, b, c. a ≠ b ∧ a ≠ c ∧ b ≠ c ∧ s = ∅ ++ a ++ b ++ c)

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

spec: set-basic inst: name: selsize seq: ⊦ # s ≠ 0 → (∃ a. a ∈ s) subst: [s] → [s] flags: current, []

Additional information

Used terms: s