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Proof step 12 from proof size-three-02 (previous step)

2 = # s0, a0 ∈ s0 ⊦ a ∈ s0, ∃ a0, b, c. a0 ≠ b ∧ a0 ≠ c ∧ b ≠ c ∧ s0 ++ a = ∅ ++ a0 ++ b ++ c

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 14:
2 = # s0, ∃ s1. s0 = s1 ++ a0 ∧ ¬ a0 ∈ s1 ⊦ a ∈ s0, ∃ a0, b, c. a0 ≠ b ∧ a0 ≠ c ∧ b ≠ c ∧ s0 ++ a = ∅ ++ a0 ++ b ++ c

Rule argument

spec: set-basic inst: name: indef seq: ⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0) subst: [s, a] → [s0, a0] flags: rewrite, current

Additional information

Used terms: s0, a0