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Proof step 5 from proof bmin-intempty (previous step)

s0.max ∈ s0, s0.min ∈ s0, ((s -- s.min) ∩ s0).min ∈ s0, s.max ∈ s, s.min ∈ s, ((s -- s.min) ∩ s0).max ∈ s, ((s -- s.min) ∩ s0).max ∈ s0, ((s -- s.min) ∩ s0).min ∈ s, ∀ a. a ∈ s ∩ s0 ↔ a ∈ ∅ ⊦ # s0 = 0, #((s -- s.min) ∩ s0) = 0, # s = 0, ((s -- s.min) ∩ s0).max = s.min, ((s -- s.min) ∩ s0).min = s.min, s = ∅, s0 = ∅, (s -- s.min) ∩ s0 = ∅, s.max < s0.min, s.max < ((s -- s.min) ∩ s0).min, s0.max < ((s -- s.min) ∩ s0).max, ((s -- s.min) ∩ s0).max < s0.min, s.max < ((s -- s.min) ∩ s0).max, s0.max < ((s -- s.min) ∩ s0).min, ((s -- s.min) ∩ s0).min < s0.min

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
((s -- s.min) ∩ s0).min ∈ s, ((s -- s.min) ∩ s0).max ∈ s0, ((s -- s.min) ∩ s0).max ∈ s, s.min ∈ s, s.max ∈ s, ((s -- s.min) ∩ s0).min ∈ s0, s0.min ∈ s0, s0.max ∈ s0, ∀ a. ¬ a ∈ s ∩ s0 ⊦ # s = 0, #((s -- s.min) ∩ s0) = 0, # s0 = 0, ((s -- s.min) ∩ s0).max = s.min, ((s -- s.min) ∩ s0).min = s.min, s = ∅, s0 = ∅, (s -- s.min) ∩ s0 = ∅, ((s -- s.min) ∩ s0).min < s0.min, s0.max < ((s -- s.min) ∩ s0).min, s.max < ((s -- s.min) ∩ s0).max, ((s -- s.min) ∩ s0).max < s0.min, s0.max < ((s -- s.min) ∩ s0).max, s.max < ((s -- s.min) ∩ s0).min, s.max < s0.min

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅; (s)