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Proof step 5 from proof size-three (previous step)

2 < # s, a ∈ s ⊦ # s = 0, ∃ 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 6:
a ∈ s, s ≠ ∅ ++ a ⊦ ∃ b. b ∈ s ∧ b ≠ a
Proof step 7:
∃ b. b ∈ s ∧ b ≠ a, 2 < # s, a ∈ s ⊦ # s = 0, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Rule argument

spec: set-basic inst: name: two-01 seq: a ∈ s, s ≠ ∅ ++ a ⊦ ∃ b. b ∈ s ∧ b ≠ a subst: [a, s] → [a, s] flags: current, []

Additional information

Used simplifier rules: size-one-01: ⊦ #(∅ ++ a) = 1; Used terms: a, s