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Proof step 6 from proof min-two-02 (previous step)

(∅ ++ a0 ++ b).min < a, (∅ ++ a0 ++ b).max = b ∨ (∅ ++ a0 ++ b).max = a0, (∅ ++ a0 ++ b).min = b ∨ (∅ ++ a0 ++ b).min = a0, a = b ∨ a = a0 ⊦ ∅ ++ a ++ (∅ ++ a0 ++ b).min = ∅ ++ a0 ++ b, a0 = b, (∅ ++ a0 ++ b).max < a, (∅ ++ a0 ++ b).max < (∅ ++ a0 ++ b).min

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: b < a0

Additional information

Used simplifier rules: min-two-01: ⊦ a < b → (∅ ++ b ++ a).min = a;%min-two: ⊦ a < b → (∅ ++ a ++ b).min = a;% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)% ⊦ ¬ a < a; (s)% ⊦ a ≠ a0 → (∅ ++ a ++ b = ∅ ++ a0 ++ c ↔ a = c ∧ b = a0); (ws)% ⊦ ∅ ++ a ++ b = ∅ ++ c ++ b ↔ a = c; (s)%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem%nen: ⊦ ¬ b < a ∧ a = c → ¬ b < c; (forward) from specification oelem Used formulas: b < a0