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Proof step 5 from proof bmin-sub-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(∅ ++ a ++ b).butmin.min ∈ s0, (∅ ++ a ++ b).min ∈ s0, s0.min ∈ s0, s0.max ∈ s0 ⊦ # s0 = 0, a = b, s0 = ∅, s0.max < s0.min, s0.max < (∅ ++ a ++ b).min, (∅ ++ a ++ b).min < s0.min, s0.max < (∅ ++ a ++ b).butmin.min, (∅ ++ a ++ b).butmin.min < s0.min, ∅ ++ a ++ b ⊆ s0

Rule argument

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;