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Proof step 6 from proof min-intempty (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: s.min abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2; (ws)%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1; from specification set-union%int-c: ⊦ s ∩ s0 = s0 ∩ s; from specification set-union Used terms: s.min