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Proof step 2 from proof min-two-02 (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 3:
⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ∧ a ∈ s ∧ b ∈ s)
Proof step 4:
# s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ∧ a ∈ s ∧ b ∈ s), # s = 2, a ∈ s, s.min ∈ s, s.max ∈ s ⊦ s = ∅, a = s.min, ∅ ++ a ++ s.min = s, s.max < a

Rule argument

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

Additional information

Used terms: s