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Proof step 2 from proof min-ins-02 (previous step)

s.min < (s ++ a).min, 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: min-q seq: a ∈ s, ∀ b. b ∈ s → ¬ b < a ⊦ a = s.min subst: [a, s] → [s.min, s ++ a] flags: rewrite, current

Additional information

Used simplifier rules: min-insert: ⊦ a < s.min ∧ s ≠ ∅ → (s ++ a).min = a;%min-ins-03: ⊦ s.min < a ∧ s ≠ ∅ → (s ++ a).min = s.min;%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%in: ⊦ a ∈ s → a ∈ s ++ b; from specification set-basic%irreflexivity: ⊦ ¬ a < a; from specification oelem%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic%transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem Used terms: s.min, s ++ a