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Proof step 6 from proof size-three-02 (previous step)

# s = 3, a ∈ s ⊦ ∃ a, b, c. a ≠ b ∧ a ≠ c ∧ b ≠ c ∧ s = ∅ ++ a ++ b ++ c

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 8:
# s = 3, ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ ∃ a, b, c. a ≠ b ∧ a ≠ c ∧ b ≠ c ∧ s = ∅ ++ a ++ b ++ c

Rule argument

spec: set-basic inst: name: indef seq: ⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0) subst: [s, a] → [s, a] flags: rewrite, current

Additional information

Used terms: s, a