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

2 < # s ⊦ # s = 0, s = ∅, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

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), 2 < # s ⊦ # s = 0, s = ∅, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ 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