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Proof step 5 from proof bmin-del-04 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: oset inst: name: bmin-del-05 seq: ⊦ s ≠ ∅ → s -- s.min = s.butmin subst: [s] → [s -- a] flags: rewrite, current

Additional information

Used simplifier rules: min-del-01: ⊦ s -- a ≠ ∅ ∧ a ≠ s.min → (s -- a).min = s.min;%bmin-del-02: ⊦ s ≠ ∅ → s -- a -- s.min = s.butmin -- a;%Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union%one-min: ⊦ s = ∅ ++ a → s -- a = ∅; from specification set-union%one-size: ⊦ s = ∅ ++ a → # s = 1; (forward) from specification set-basic%in-06: ⊦ a ∈ s ∧ s = ∅ ++ b → a = b; (forward) from specification set-basic Used terms: s -- a