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Proof step 5 from proof bmin-two-08 (previous step)

⊦ ∅ ++ (∅ ++ a ++ b).butmin.min ++ (∅ ++ a ++ b).min = ∅ ++ a ++ b, a = b

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;%sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%eq-04: ⊦ s ++ b ++ a = s ++ a ++ b ↔ true; from specification set-basic%eqtwo-01: ⊦ ∅ ++ a ++ b = ∅ ++ c ++ b ↔ a = c; from specification set-basic Used formulas: b < a