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Proof step 1 from proof min-intempty (conclusion)

⊦ s ∩ s0 = ∅ ∧ s ≠ ∅ → (¬ ¬ s.min ∈ s0 ↔ false)

Heuristic simplifier applied rule simplifier

And closed the goal!

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;%int-empty-02: ⊦ s ∩ s0 = ∅ ∧ a ∈ s → (¬ ¬ a ∈ s0 ↔ false); from specification set-union%int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic