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Proof step 11 from proof rely-owns-none (previous step)

f(r) = f0(r), g0(r) = g(r), r ∈ rs0, r ∈ rs, ∀ 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, # rs0 = 0, rs = ∅, rs0 = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: r

Additional information

Used terms: r