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Proof step 2 from proof bmax-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s ≠ ∅ → s.butmax = s -- s.max

Rule argument

spec: oset inst: name: Butmax seq: ⊦ s ≠ ∅ → s.butmax = s -- s.max subst: [s] → [∅ ++ a] flags: rewrite, current

Additional information

Used simplifier rules: max-one: ⊦ (∅ ++ a).max = a;%del-addeq-05: ⊦ (s ++ a) -- a = s -- a; from specification set-union%del-e-01: ⊦ ∅ -- a = ∅; from specification set-union%ne: ⊦ ∅ ≠ s ++ a; from specification set-basic Used terms: ∅ ++ a