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Proof step 1 from proof max-q (conclusion)

a ∈ s, ∀ b. b ∈ s → ¬ a < b ⊦ s.max = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a < s.max, s.min < s.max, a ∈ s, s.min ∈ s, s.max ∈ s, ∀ b. b ∈ s → ¬ a < b ⊦ # s = 0, s = ∅, s.max < s.min, a < s.min

Rule argument

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;% ⊦ ¬ a < a; (s)%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic% ⊦ a ≠ b → (a < b ↔ ¬ b < a); (ss)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic