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Proof step 2 from proof max-del-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
a ∈ s, ∀ b. b ∈ s → ¬ a < b ⊦ s.max = a

Rule argument

spec: oset inst: name: max-q seq: a ∈ s, ∀ b. b ∈ s → ¬ a < b ⊦ s.max = a subst: [a, s] → [s.max, s -- a] flags: rewrite, current

Additional information

Used simplifier rules: max-in: ⊦ s ≠ ∅ → (¬ s.max ∈ s ↔ false);%max-gr: ⊦ s.max < a ∧ s ≠ ∅ → ¬ a ∈ s;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s; from specification set-union%del-rax: ⊦ a ∈ s -- b ↔ ¬ ¬(a ≠ b ∧ a ∈ s); from specification set-union Used terms: s.max, s -- a