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Proof step 3 from proof rely-owned-unchgd (previous step)

∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (a[P] * true)[rs, f, g] ↔ (a[P] * true)[rs0, f0, g0]

User applied rule case distinction

And obtained the following new goals

Proof step 4:
(a[P] * true)[rs, f, g], ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (a[P] * true)[rs0, f0, g0]
Proof step 47:
(a[P] * true)[rs0, f0, g0], ∀ r. (g(r) = a ∧ r ∈ rs ↔ g0(r) = a ∧ r ∈ rs0) ∧ (g(r) = a ∧ r ∈ rs → f(r) = f0(r)) ⊦ (a[P] * true)[rs, f, g]

Rule argument

pos: right 1

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