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Proof step 6 from proof size-three (previous step)

2 < # s, a ∈ 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 7:
a ∈ s, s ≠ ∅ ++ a ⊦ ∃ b. b ∈ s ∧ b ≠ a
Proof step 8:
2 < # s, a ∈ s, ∃ b. b ∈ s ∧ b ≠ a ⊦ ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Rule argument

spec: <toplevel> name: two-01 seq: a ∈ s, s ≠ ∅ ++ a ⊦ ∃ b. b ∈ s ∧ b ≠ a subst: [a, s] → [a, s]

Additional information

Used simplifier rules: in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;%size-one-01: ⊦ #(∅ ++ a) = 1; Used terms: a, s