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Proof step 9 from proof rem-inter-empty3 (previous step)

(s \ s0) ∩ s1 = ∅, a ∈ s2, a ∈ s1, ∀ a0. ¬ a0 ∈ s2 \ s ∨ ¬ a0 ∈ s1 ⊦ #((s2 \ s0) ∩ s1) = 0, # s2 = 0, # s1 = 0, s1 = ∅, s2 = ∅, a ∈ s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 11:
a ∈ s1, a ∈ s2, ∀ a0. ¬ a0 ∈ s2 \ s ∨ ¬ a0 ∈ s1, ∀ a. a ∈ (s \ s0) ∩ s1 ↔ a ∈ ∅ ⊦ # s1 = 0, # s2 = 0, #((s2 \ s0) ∩ s1) = 0, s1 = ∅, s2 = ∅, a ∈ s0

Rule argument

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

Additional information

Used terms: ∅, (s \ s0) ∩ s1