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Proof step 2 from proof rely-owns-one (previous step)

rely(a, rs, f, g, rs0, f0, g0) ⊦ owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ owns-one(a, r, rs, g) ↔ owns-one(a, r, rs0, g0)

Rule argument

Additional information

Used simplifier rules: rely: ⊦ rely(a, rs0, f0, g0, rs1, f1, g1) ↔ (∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)));