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Proof step 2 from proof maxunion (previous step)

(s ∪ s0).max < s.max, s.min ∈ s, s.max ∈ s ⊦ s = ∅, # s = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: oset inst: name: Max-max seq: ⊦ a ∈ s → ¬ s.max < a subst: [a, s] → [s.max, s ∪ s0] flags: rewrite, current

Additional information

Used simplifier rules: max-in: ⊦ s ≠ ∅ → (¬ s.max ∈ s ↔ false);%unin-01: ⊦ ¬ a ∈ s ∪ s0 ↔ ¬ a ∈ s ∧ ¬ a ∈ s0; from specification set-union%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1; from specification set-union%unc: ⊦ s ∪ s0 = s0 ∪ s; from specification set-union Used terms: s.max, s ∪ s0