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Proof step 3 from proof min-intempty (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 5:
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

Rule argument

spec: set-basic name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [∅, s ∩ s0] flags: <all paths>

Additional information

Used terms: ∅, s ∩ s0