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Proof step 3 from proof bmin-sub-02 (previous step)

# s = 2, s.butmin.min ∈ s0, s.min ∈ s0, s0.min ∈ s0, s0.max ∈ s0 ⊦ # s0 = 0, s0 = ∅, s.min < s0.min, s0.max < s.min, s.butmin.min < s0.min, s0.max < s.butmin.min, s0.max < s0.min, s ⊆ s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b)
Proof step 5:
s0.max ∈ s0, s0.min ∈ s0, s.min ∈ s0, s.butmin.min ∈ s0, ∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b ⊦ # s0 = 0, s0 = ∅, s0.max < s0.min, s0.max < s.butmin.min, s.butmin.min < s0.min, s0.max < s.min, s.min < s0.min, s ⊆ s0

Rule argument

spec: <toplevel> name: size-two seq: ⊦ # s = 2 ↔ (∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b) subst: [s] → [s] flags: <all paths>

Additional information

Used terms: s