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Proof step 32 from proof del-lists-unions (previous step)

P(H1), del-list(rx0, n0)(H), ls(r, x)(H2), H2 ⊥ H, H1 ⊥ H2, H1 ⊥ H, ∀ n, H0, H1, m, P. del-list(rx0, n)(H0) ∧ H0 ⊥ H1 → (((del-list(rx0, m) * P))(H0 ∪ H1) ↔ n = m ∧ P(H1)) ⊦ r = null, (((ls(r, x) * del-list(rx0, n0)) * P))((H2 ∪ H) ∪ H1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 33:
⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H
Proof step 34:
P(H1), del-list(rx0, n0)(H), ls(r, x)(H2), H1 ⊥ H, H2 ⊥ H, H ⊥ H1, H1 ⊥ H2, ∀ n, H0, H1, m, P. del-list(rx0, n)(H0) ∧ H0 ⊥ H1 → (((del-list(rx0, m) * P))(H0 ∪ H1) ↔ n = m ∧ P(H1)) ⊦ r = null, (((ls(r, x) * del-list(rx0, n0)) * P))(H2 ∪ H ∪ H1)

Rule argument

spec: heap, inst: cell-heap name: uni-ass seq: ⊦ H0 ⊥ H1 ∧ H1 ⊥ H ∧ H ⊥ H0 → (H0 ∪ H1) ∪ H = H0 ∪ H1 ∪ H subst: [H, H1, H0] → [H1, H, H2] flags: <all paths>

Additional information

Used simplifier rules: ⊦ H0 ⊥ H1 → H1 ⊥ H0; (s) Used terms: H1, H, H2