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Proof step 4 from proof disj (previous step)

∀ r. ¬ r ∈ rs0 ∨ ¬ r ∈ rs1 ⊦ #(rs0 ∩ rs1) = 0, rs0 ∩ rs1 = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ rs = rs0 ↔ (∀ r. r ∈ rs ↔ r ∈ rs0)
Proof step 6:
∀ r. ¬ r ∈ rs0 ∨ ¬ r ∈ rs1 ⊦ #(rs0 ∩ rs1) = 0, ∀ r. r ∈ rs0 ∩ rs1 ↔ r ∈ ∅

Rule argument

spec: set-basic, inst: refset name: Extension seq: ⊦ rs = rs0 ↔ (∀ r. r ∈ rs ↔ r ∈ rs0) subst: [rs0, rs] → [∅, rs0 ∩ rs1] flags: <all paths>

Additional information

Used terms: ∅, rs0 ∩ rs1