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Proof step 3 from proof min-two-02 (previous step)

# s = 2, s.min < a, a ∈ s, s.min ∈ s, s.max ∈ s ⊦ ∅ ++ a ++ s.min = s, s = ∅, s.max < s.min, s.max < a

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ∧ a ∈ s ∧ b ∈ s)
Proof step 5:
# s = 2, # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ∧ a ∈ s ∧ b ∈ s), s.min < a, s.max ∈ s, s.min ∈ s, a ∈ s ⊦ ∅ ++ a ++ s.min = s, s = ∅, s.max < a, s.max < s.min

Rule argument

spec: set-basic name: size-two-05 seq: ⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ∧ a ∈ s ∧ b ∈ s) subst: [s] → [s]

Additional information

Used terms: s