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Proof step 8 from proof size-three (previous step)

b ∈ s, 2 < # s, a ∈ s ⊦ # s = 0, b = a, ∃ 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 9:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s
Proof step 10:
b ∈ s, 2 < # s, a ∈ s ⊦ #(∅ ++ a ++ b) < # s, # s = 0, b = a, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c
Proof step 12:
∃ a0. a0 ∈ s ∧ ¬ a0 ∈ ∅ ++ a ++ b, b ∈ s, 2 < # s, a ∈ s ⊦ # s = 0, b = a, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Rule argument

spec: set-basic inst: name: size-exgr seq: # s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s subst: [s, s0] → [∅ ++ a ++ b, s] flags: current, []

Additional information

Used terms: ∅ ++ a ++ b, s