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Proof step 6 from proof bmin-sub-02 (previous step)

(∅ ++ 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

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;%min-two-01: ⊦ a < b → (∅ ++ b ++ a).min = a;%bmin-two-01: ⊦ a < b → (∅ ++ a ++ b).butmin = ∅ ++ b;%min-one: ⊦ (∅ ++ a).min = a;%min-two: ⊦ a < b → (∅ ++ a ++ b).min = a;% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)% ⊦ a ∈ s0 → (s ++ a ⊆ s0 ↔ s ⊆ s0); (s)% ⊦ ∅ ⊆ s; (s) Used formulas: b < a