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Proof step 9 from proof size-three-02 (previous step)

2 = # s0 ⊦ a ∈ s0, ∃ a0, b, c. a0 ≠ b ∧ a0 ≠ c ∧ b ≠ c ∧ s0 ++ a = ∅ ++ a0 ++ b ++ c

User applied rule insert lemma

And obtained the following new goals

Proof step 10:
⊦ # s ≠ 0 → (∃ a. a ∈ s)
Proof step 11:
# s0 ≠ 0 → (∃ a. a ∈ s0), 2 = # s0 ⊦ a ∈ s0, ∃ a0, b, c. a0 ≠ b ∧ a0 ≠ c ∧ b ≠ c ∧ s0 ++ a = ∅ ++ a0 ++ b ++ c

Rule argument

spec: set-basic inst: name: selsize seq: ⊦ # s ≠ 0 → (∃ a. a ∈ s) subst: [s] → [s0] flags: current, []

Additional information

Used terms: s0