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Proof step 1 from proof bmin-in-04 (conclusion)

s.butmin ≠ ∅ ⊦ s ⊆ s0 ∧ s ≠ ∅ → s.butmin.min ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
s.max ∈ s, s.min ∈ s, s ⊆ s0, s.max ∈ s0, s.min ∈ s0, s0.min ∈ s0, s0.max ∈ s0 ⊦ s0 = ∅, # s = 0, s = ∅ ++ s.min, s = ∅, # s0 = 0, s.min < s0.min, s0.max < s.min, s.max < s0.min, s0.max < s.max, (s -- s.min).min ∈ s0

Rule argument

Additional information

Used simplifier rules: Butmin: ⊦ s ≠ ∅ → s.butmin = s -- s.min;%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;%nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem%irreflexivity: ⊦ ¬ a < a; from specification oelem%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a; from specification set-union