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Proof step 10 from proof rely-owns-none (previous step)

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 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
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 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ rs ≠ ∅ → # rs ≠ 0; (forward) from specification set-basic[refset]%ne-01: ⊦ r ∈ rs → rs ≠ ∅; (forward) from specification set-basic[refset]