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Proof step 6 from proof bmin-two-04 (previous step)

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