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Proof step 5 from proof min-two-02 (previous step)

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

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;%sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem%ls-le: ⊦ a < b → (¬ b < a ↔ true); from specification oelem%two: ⊦ a ≠ a0 → (∅ ++ a ++ b = ∅ ++ a0 ++ c ↔ a = c ∧ b = a0); from specification set-basic%eqtwo-01: ⊦ ∅ ++ a ++ b = ∅ ++ c ++ b ↔ a = c; from specification set-basic Used formulas: b < a0