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Proof step 2 from proof butmin-ins-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: oset inst: name: Butmin seq: ⊦ s ≠ ∅ → s.butmin = s -- s.min subst: [s] → [s ++ a] flags: rewrite, current

Additional information

Used simplifier rules: min-insert: ⊦ a < s.min ∧ s ≠ ∅ → (s ++ a).min = a;%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%del-addeq-05: ⊦ (s ++ a) -- a = s -- a; from specification set-union%del-in-03: ⊦ s -- a = s ↔ ¬ a ∈ s; from specification set-union%eq: ⊦ a ∈ s → s ++ a = s; from specification set-basic%ne: ⊦ ∅ ≠ s ++ a; from specification set-basic Used terms: s ++ a