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Proof step 2 from proof rely-alloc (previous step)

⊦ a0 = a, rely(a, rs, f, g, rs ++ r, f, g(r; a0)), r ∈ rs

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: rely: ⊦ 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)));% ⊦ r ∈ rs ++ r0 ↔ r = r0 ∨ r ∈ rs; (s)% ⊦ r ∈ rs → r ∈ rs ++ r0; (s)