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

2 < # s, b ∈ s, a ∈ s ⊦ 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 10:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s
Proof step 11:
2 < # s, a ∈ s, b ∈ s ⊦ b = a, #(∅ ++ a ++ b) < # s, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c
Proof step 13:
2 < # s, a ∈ s, b ∈ s, ∃ a0. a0 ∈ s ∧ ¬ a0 ∈ ∅ ++ a ++ b ⊦ b = a, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

Rule argument

spec: <toplevel> name: size-exgr seq: # s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s subst: [s, s0] → [∅ ++ a ++ b, s]

Additional information

Used terms: ∅ ++ a ++ b, s