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Proof step 17 from proof size-three-02 (previous step)

1 = # s ⊦ a = a0, a0 ∈ s, a ∈ s, ∃ a1, b, c. a1 ≠ b ∧ a1 ≠ c ∧ b ≠ c ∧ s ++ a0 ++ a = ∅ ++ a1 ++ b ++ c

User applied rule insert lemma

And obtained the following new goals

Proof step 18:
⊦ # s ≠ 0 → (∃ a. a ∈ s)
Proof step 19:
# s ≠ 0 → (∃ a. a ∈ s), 1 = # s ⊦ a = a0, a0 ∈ s, a ∈ s, ∃ a1, b, c. a1 ≠ b ∧ a1 ≠ c ∧ b ≠ c ∧ s ++ a0 ++ a = ∅ ++ a1 ++ 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