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

s.max < s.butmax.max, 1 < # s ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Butmax: ⊦ s ≠ ∅ → s.butmax = s -- s.max;%max-del: ⊦ s.max < (s -- a).max → s = ∅ ++ a;%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;%in-06: ⊦ a ∈ s ∧ s = ∅ ++ b → a = b; (forward) from specification set-basic%one-size: ⊦ s = ∅ ++ a → # s = 1; (forward) from specification set-basic% ⊦ s = ∅ ++ a → s -- a = ∅; (s)%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)