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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅; (s)