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Proof step 7 from proof min-insert (previous step)

a < s.min, s.max ∈ s, s.min ∈ s ⊦ # s = 0, (s ++ a).min = a, s = ∅, (s ++ a).min < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a < s.min, a < (s ++ a).min, s.min ∈ s, s.max ∈ s ⊦ # s = 0, (s ++ a).min = a, s = ∅

Rule argument

Additional information

Used simplifier rules: ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)