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Proof step 3 from proof rely-owns-one (previous step)

∀ 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)

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0, g(r) = a ∧ r ∈ rs → f(r) = f0(r), ∀ 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

pos: left 1 terms: r

Additional information

Used terms: r