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Proof step 6 from proof alloc (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ (r ==>_ * R)[rs, f, g] ↔ (∃ a. (a[r =>_] * R)[rs, f, g])
Proof step 8:
⊦ (∃ a. (a[r =>_] * R)[rs ++ r, f, g]) ↔ (a[r =>_] * R)[rs ++ r, f, g(r; a)], r = null, r ∈ rs

Rule argument

spec: <toplevel> name: lift-unknown seq: ⊦ (r ==>_ * R)[rs, f, g] ↔ (∃ a. (a[r =>_] * R)[rs, f, g]) subst: [g, f, rs, R, r] → [g, f, rs ++ r, R, r] flags: <all paths>

Additional information

Used terms: g, f, rs ++ r, R, r