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Proof step 9 from proof bmin-sub-01 (previous step)

a ∈ s0, b ∈ s0, s0.min ∈ s0, s0.max ∈ s0 ⊦ # s0 = 0, a = b, s0 = ∅, b < s0.min, s0.max < b, a < s0.min, s0.max < a, s0.max < s0.min, (∅ ++ a ++ b).butmin.min ∈ s0

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: b < a

Additional information

Used simplifier rules: bmin-two-02: ⊦ a < b → (∅ ++ b ++ a).butmin = ∅ ++ b;%min-one: ⊦ (∅ ++ a).min = a;%bmin-two-01: ⊦ a < b → (∅ ++ a ++ b).butmin = ∅ ++ b;%min-one: ⊦ (∅ ++ a).min = a;% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss) Used formulas: b < a