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Proof step 19 from proof rely-owned-unchgd (previous step)

rs2 ∩ rs1 = ∅, P(mk(rs2, f)), ∀ r. r ∈ rs2 → g(r) = a, ∀ r. (g(r) = a ∧ r ∈ rs2 ∪ rs1 ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs2 ∪ rs1 → f(r) = f0(r)) ⊦ rs2 ∪ (rs0 \ rs2) = rs0, rs2 ⊆ rs0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ rs ⊆ rs0 ↔ (∀ r. r ∈ rs → r ∈ rs0)
Proof step 21:
rs2 ∩ rs1 = ∅, P(mk(rs2, f)), ∀ r. (g(r) = a ∧ r ∈ rs2 ∪ rs1 ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs2 ∪ rs1 → f(r) = f0(r)), ∀ r. r ∈ rs2 → g(r) = a ⊦ rs2 ∪ (rs0 \ rs2) = rs0, ∀ r. r ∈ rs2 → r ∈ rs0

Rule argument

spec: set-basic, inst: refset name: Subset seq: ⊦ rs ⊆ rs0 ↔ (∀ r. r ∈ rs → r ∈ rs0) subst: [rs0, rs] → [rs0, rs2] flags: <all paths>

Additional information

Used terms: rs0, rs2