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Proof step 8 from proof alloc (previous step)

⊦ (∃ a. (a[r =>_] * R)[rs ++ r, f, g]) ↔ (a[r =>_] * R)[rs ++ r, f, g(r; a)], r = null, r ∈ rs

User applied rule case distinction

And obtained the following new goals

Proof step 9:
∃ a. (a[r =>_] * R)[rs ++ r, f, g] ⊦ r = null, (a[r =>_] * R)[rs ++ r, f, g(r; a)], r ∈ rs
Proof step 12:
(a[r =>_] * R)[rs ++ r, f, g(r; a)] ⊦ r = null, r ∈ rs, ∃ a. (a[r =>_] * R)[rs ++ r, f, g]

Rule argument

pos: right 1

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