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Proof step 2 from proof max-ifins (previous step)

s ≠ ∅ ⊦ (s ++ a).max = (a < s.max ⊃ s.max;a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
s.min ∈ s, s.max ∈ s ⊦ (s ++ a).max = (a < s.max ⊃ s.max;a), s = ∅, # s = 0

Rule argument

Additional information

Used simplifier rules: Max-in: ⊦ s ≠ ∅ → s.max ∈ s;%Min-in: ⊦ s ≠ ∅ → s.min ∈ s;%Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%minmax-04: ⊦ s ≠ ∅ → ¬ s.max < s.min;%irreflexivity: ⊦ ¬ a < a; from specification oelem%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic