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Proof step 13 from proof alloc (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ (a0[r =>_] * R)[rs, f, g] ↔ (a1[r =>_] * R)[rs, f, g(r; a1)] ∧ g(r) = a0

Rule argument

spec: <toplevel> name: aux-unknown seq: ⊦ (a0[r =>_] * R)[rs, f, g] ↔ (a1[r =>_] * R)[rs, f, g(r; a1)] ∧ g(r) = a0 subst: [a1, g, f, rs, R, r, a0] → [a, g, f, rs ++ r, R, r, g(r)] flags: <all paths>

Additional information

Used terms: a, g, f, rs ++ r, R, r, g(r)