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Proof step 3 from proof alloc (previous step)

rely(me, O1(r; me), H1 ++ r, O2, H2) ⊦ ((P0 * true))(H1) ↔ ((!(me, O2, r =>_) * P0 * true))(H1 ++ r), r = null, r ∈ H1

User applied rule apply lemma

And obtained the following new goals

Proof step 4:
⊦ r ≠ null ∧ ¬ r ∈ H1 ∧ rely(me, O1(r; me), H1 ++ r, O2, H2) → (((!(me, O1, P) * P0))(H1) ↔ ((!(me, O2, r =>_ * P) * P0))(H1 ++ r))

Rule argument

spec: <toplevel> name: alloc-p seq: ⊦ r ≠ null ∧ ¬ r ∈ H1 ∧ rely(me, O1(r; me), H1 ++ r, O2, H2) → ( ((!(me, O1, P) * P0))(H1) ↔ ((!(me, O2, r =>_ * P) * P0))(H1 ++ r)) subst: [r, H1, me, O1, O2, H2, P, P0] → [r, H1, me, O1, O2, H2, emp, P0 * true]

Additional information

Used simplifier rules: owned-emp: ⊦ !(me, O, emp) = emp;% ⊦ emp * P = P; (s)%sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep Used terms: r, H1, me, O1, O2, H2, emp, P0 * true