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Proof step 21 from proof owner-lst1 (previous step)

mk(rs1, cf) = H0 ∪ H, rs1 ∩ rs0 = ∅, ((r => a × r0))(H0), R[rs0, cf, af], ls(r0, x0)(H), H0 ⊥ H ⊦ af(r) = o, r = null, r ∈ rs1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r]))
Proof step 23:
rs1 ∩ rs0 = ∅, ((r => a × r0))(H0), R[rs0, cf, af], ls(r0, x0)(H), H0 ⊥ H, ∀ r. (r ∈ mk(rs1, cf) ↔ r ∈ H0 ∪ H) ∧ (r ∈ mk(rs1, cf) → mk(rs1, cf)[r] = (H0 ∪ H)[r]) ⊦ af(r) = o, r = null, r ∈ rs1

Rule argument

spec: heap, inst: cell-heap-aux name: ext seq: ⊦ H0 = H1 ↔ (∀ r. (r ∈ H0 ↔ r ∈ H1) ∧ (r ∈ H0 → H0[r] = H1[r])) subst: [H1, H0] → [H0 ∪ H, mk(rs1, cf)] flags: <all paths>

Additional information

Used terms: H0 ∪ H, mk(rs1, cf)