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Proof step 22 from proof rely-owned-unchgd (previous step)

rs2 ∩ rs1 = ∅, P(mk(rs2, f)), r ∈ rs2, ∀ 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 = 0, rs2 = ∅, rs2 ∪ (rs0 \ rs2) = rs0, r ∈ rs0

User applied rule all left

And obtained the following new goals

Proof step 23:
g(r) = a ∧ r ∈ rs2 ∪ rs1 ↔ g0(r) = a ∧ r ∈ rs0, rs2 ∩ rs1 = ∅, P(mk(rs2, f)), r ∈ rs2, g(r) = a ∧ r ∈ rs2 ∪ rs1 → f(r) = f0(r), ∀ r. r ∈ rs2 → g(r) = a ⊦ # rs2 = 0, rs2 = ∅, rs2 ∪ (rs0 \ rs2) = rs0, r ∈ rs0

Rule argument

pos: left 4 terms: r abort: F discard: T computed: T

Additional information

Used terms: r