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Proof step 7 from proof size-two (previous step)

a ∈ s ⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0)
Proof step 9:
∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b)

Rule argument

spec: <toplevel> name: indef seq: ⊦ a ∈ s ↔ (∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0) subst: [s, a] → [s, a] flags: <all paths>

Additional information

Used terms: s, a