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Proof step 10 from proof rely-owns-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
f(r) = f0(r), g(r) = g0(r0), g0(r) = g0(r0), r0 ∈ rs0, r ∈ rs0, r ∈ rs, ∀ r. (g(r) = g0(r0) ∧ r ∈ rs ↔ g0(r) = g0(r0) ∧ r ∈ rs0) ∧ (g(r) = g0(r0) ∧ r ∈ rs → f(r) = f0(r)), ∀ r0. r0 ∈ rs ∧ r0 ≠ r → g(r0) ≠ g(r) ⊦ # rs0 = 0, # rs = 0, r = null, r0 = r, rs = ∅, rs0 = ∅

Proof step 10 from proof rely-owns-one (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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