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Proof step 5 from proof bm-shift (previous step)

s.min ∈ s, s.max ∈ s ⊦ s = ∅, # s = 0, # s = 1, s -- s.min -- (s -- s.min).max = (s -- s.max).butmin, s = ∅ ++ s.min

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ s ≠ ∅ → s.butmin = s -- s.min
Proof step 7:
s -- s.max ≠ ∅, s.min ∈ s, s.max ∈ s ⊦ s = ∅, # s = 0, # s = 1, s -- s.min -- (s -- s.min).max = s -- s.max -- (s -- s.max).min, s = ∅ ++ s.min

Rule argument

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

Additional information

Used simplifier rules: 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 -- s.max