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Proof step 4 from proof max-ifins (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
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-gr: ⊦ s.max < a ∧ s ≠ ∅ → ¬ a ∈ s;%in: ⊦ a ∈ s → a ∈ s ++ b; from specification set-basic%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem%lel: ⊦ a < b ∧ a = c → c < b; (forward) from specification oelem%irreflexivity: ⊦ ¬ a < a; from specification oelem Used terms: s.max, s ++ a