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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ s ≠ ∅ ∧ a ≠ s.min → (s -- a).min = s.min

Rule argument

spec: oset inst: name: min-del seq: ⊦ s ≠ ∅ ∧ a ≠ s.min → (s -- a).min = s.min subst: [a, s] → [s.max, s] flags: rewrite, current

Additional information

Used simplifier rules: minmax-02: ⊦ s.min = s.max ∧ s ≠ ∅ → # s = 1; Used terms: s.max, s