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Proof step 7 from proof bmin-sub-01 (previous step)

∃ a, b. a ≠ b ∧ s = ∅ ++ a ++ b, s ⊆ s0 ⊦ s.butmin.min ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
b ∈ s0, a ∈ s0, s0.min ∈ s0, s0.max ∈ s0 ⊦ s0 = ∅, a = b, # s0 = 0, a < s0.min, s0.max < a, b < s0.min, s0.max < b, (∅ ++ a ++ b).butmin.min ∈ s0

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;%minmax-04: ⊦ s ≠ ∅ → ¬ s.max < s.min;%irreflexivity: ⊦ ¬ a < a; from specification oelem%nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%sub-one-02: ⊦ ∅ ++ a ⊆ s ↔ a ∈ s; from specification set-basic%sub-add-03: ⊦ s ++ a ⊆ s0 ↔ s ⊆ s0 ∧ a ∈ s0; from specification set-basic