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Proof step 18 from proof sep (previous step)

rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), ∀ r. (r ∈ rs ↔ r ∈ rs0 ∨ r ∈ rs1) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦ rs0 ∪ rs1 = rs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ rs = rs0 ↔ (∀ r. r ∈ rs ↔ r ∈ rs0)
Proof step 20:
rs0 ∩ rs1 = ∅, P0(mk(rs0, f0)), P1(mk(rs1, f1)), ∀ r. (r ∈ rs ↔ r ∈ rs0 ∨ r ∈ rs1) ∧ (r ∈ rs → f(r) = (mk(rs0, f0) ∪ mk(rs1, f1))[r]) ⊦ ∀ r. r ∈ rs0 ∪ rs1 ↔ r ∈ rs

Rule argument

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

Additional information

Used terms: rs, rs0 ∪ rs1