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Proof step 3 from proof rely-owns-none (previous step)

∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (∀ r. r ∈ rs → g(r) ≠ a) ↔ (∀ r. r ∈ rs0 → g0(r) ≠ a)

User applied rule case distinction

And obtained the following new goals

Proof step 4:
∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)), ∀ r. r ∈ rs → g(r) ≠ a ⊦ ∀ r. r ∈ rs0 → g0(r) ≠ a
Proof step 8:
∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)), ∀ r. r ∈ rs0 → g0(r) ≠ a ⊦ ∀ r. r ∈ rs → g(r) ≠ a

Rule argument

pos: right 1

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