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Proof step 2 from proof rely-refl (
previous step
)
⊦ rely(a, rs0, f0, g0, rs1, f1, g1) ↔ (∀ r. (g0(r) = a ∧ r ∈ rs0 ↔ g1(r) = a ∧ r ∈ rs1) ∧ (g0(r) = a ∧ r ∈ rs0 → f0(r) = f1(r)))
Lemma
rely