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Proof step 53 from proof rely-owned-unchgd (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 54:
rs2 ∩ rs1 = ∅, a[P][rs2, f0, g0], ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs2 ∪ rs1) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ rs2 ∪ (rs \ rs2) = rs ∧ rs2 ∩ (rs \ rs2) = ∅ ∧ a[P][rs2, f, g]

Rule argument

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

Additional information

Used terms: rs2, rs \ rs2