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Proof step 2 from proof min-intempty (previous step)

s ∩ s0 = ∅ ⊦ s.min ∈ s0 ↔ false, s = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
s ∩ s0 = ∅, s.max ∈ s, s.min ∈ s, s.min ∈ s0, s0.min ∈ s0, s0.max ∈ s0 ⊦ # s = 0, # s0 = 0, s = ∅, s0 = ∅, s.min < s0.min, s0.max < s.min

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;% ⊦ ¬ a < a; (s)%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