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Proof step 6 from proof min-insert (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a < s.min, s.min ∈ s, s.max ∈ s, a < (s ++ a).min ⊦ (s ++ a).min = a, s = ∅, # s = 0

Rule argument

Additional information

Used simplifier rules: sls: ⊦ a ≠ b → (¬ a < b ↔ b < a); from specification oelem