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Proof step 2 from proof bmin-min (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: oset inst: name: Min-min seq: ⊦ a ∈ s → ¬ a < s.min subst: [s, a] → [s, (s -- s.min).min] flags: rewrite, current

Additional information

Used simplifier rules: min-delmin-01: ⊦ s ≠ ∅ ∧ s ≠ ∅ ++ s.min → (s -- s.min).min ∈ s; Used terms: s, (s -- s.min).min