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Proof step 3 from proof owner-ls1 (previous step)

(o[ls(r, x)] * R)[rs, cf, af] ⊦ af(r) = o, r = null

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ (R0 * R1)[rs, cf, af] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ R0[rs0, cf, af] ∧ R1[rs1, cf, af])
Proof step 5:
∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ o[ls(r, x)][rs0, cf, af] ∧ R[rs1, cf, af] ⊦ af(r) = o, r = null

Rule argument

spec: heap-fun, inst: cell-heap-aux-heap-aux name: sep seq: ⊦ (R0 * R1)[rs, cf, af] ↔ (∃ rs0, rs1. rs0 ∪ rs1 = rs ∧ rs0 ∩ rs1 = ∅ ∧ R0[rs0, cf, af] ∧ R1[rs1, cf, af]) subst: [af, cf, rs, R1, R0] → [af, cf, rs, R, o[ls(r, x)]] flags: <all paths>

Additional information

Used terms: af, cf, rs, R, o[ls(r, x)]