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Proof step 2 from proof size-two-05 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b)

Rule argument

spec: set-basic inst: name: size-two seq: ⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b) subst: [s] → [s] flags: rewrite, current

Additional information

Used terms: s