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Proof step 5 from proof rem-inter-empty3 (previous step)

(s0 \ s1) ∩ s = ∅, (s2 \ s0) ∩ s = ∅, a ∈ s2, a ∈ s ⊦ s2 = ∅, s = ∅, #((s2 \ s1) ∩ s) = 0, # s2 = 0, # s = 0, a ∈ s1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 7:
(s0 \ s1) ∩ s = ∅, ∀ a. a ∈ (s2 \ s0) ∩ s ↔ a ∈ ∅, a ∈ s2, a ∈ s ⊦ s2 = ∅, s = ∅, #((s2 \ s1) ∩ s) = 0, # s2 = 0, # s = 0, a ∈ s1

Rule argument

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

Additional information

Used terms: ∅, (s2 \ s0) ∩ s