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Proof step 10 from proof rely-owned-unchgd (previous step)

rs2 ∩ rs1 = ∅, a[P][rs2, f, g], ∀ r. (g(r) = a ∧ r ∈ rs2 ∪ rs1 ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs2 ∪ rs1 → f(r) = f0(r)) ⊦ ∃ rs, rs3. rs ∪ rs3 = rs0 ∧ rs ∩ rs3 = ∅ ∧ a[P][rs, f0, g0]

User applied rule exists right

And obtained the following new goals

Proof step 11:
rs2 ∩ rs1 = ∅, a[P][rs2, f, g], ∀ 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 \ rs2) = ∅ ∧ a[P][rs2, f0, g0]

Rule argument

pos: right 1 terms: rs2, rs0 \ rs2 abort: F discard: T computed: F

Additional information

Used terms: rs2, rs0 \ rs2