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Proof step 3 from proof maxunion (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ∈ s → ¬ s.max < a

Rule argument

spec: <toplevel> name: Max-max seq: ⊦ a ∈ s → ¬ s.max < a subst: [a, s] → [s.max, s ∪ s0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2; (ws)%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