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Proof step 9 from proof rely-owns-none (previous step)

r ∈ rs, ∀ r0. (g(r0) = g(r) ∧ r0 ∈ rs ↔ g0(r0) = g(r) ∧ r0 ∈ rs0) ∧ (g(r0) = g(r) ∧ r0 ∈ rs → f(r0) = f0(r0)), ∀ r0. r0 ∈ rs0 → g0(r0) ≠ g(r) ⊦ # rs = 0, rs = ∅

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 10:
g(r) = g(r) ∧ r ∈ rs ↔ g0(r) = g(r) ∧ r ∈ rs0, r ∈ rs, g(r) = g(r) ∧ r ∈ rs → f(r) = f0(r), ∀ r0. r0 ∈ rs0 → g0(r0) ≠ g(r), ∀ r0. (g(r0) = g(r) ∧ r0 ∈ rs ↔ g0(r0) = g(r) ∧ r0 ∈ rs0) ∧ (g(r0) = g(r) ∧ r0 ∈ rs → f(r0) = f0(r0)) ⊦ # rs = 0, rs = ∅

Rule argument

pos: left 2 terms: r

Additional information

Used terms: r