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Proof step 12 from proof del-lists-unions (previous step)

((ls(r, x) * del-list(rx0, n0)))(H0), H0 ⊥ H1, ∀ n, H0, H1, m, P. del-list(rx0, n)(H0) ∧ H0 ⊥ H1 → (((del-list(rx0, m) * P))(H0 ∪ H1) ↔ n = m ∧ P(H1)) ⊦ (∃ x0. # x0 ≤ m ∧ (((ls(r, x0) * del-list(rx0, m - # x0)) * P))(H0 ∪ H1)) ↔ n0 + # x = m ∧ P(H1), r = null

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1))
Proof step 14:
H0 ⊥ H1, ∀ n, H0, H1, m, P. del-list(rx0, n)(H0) ∧ H0 ⊥ H1 → (((del-list(rx0, m) * P))(H0 ∪ H1) ↔ n = m ∧ P(H1)), ∃ H2, H1. H2 ⊥ H1 ∧ H0 = H2 ∪ H1 ∧ ls(r, x)(H2) ∧ del-list(rx0, n0)(H1) ⊦ (∃ x0. # x0 ≤ m ∧ (((ls(r, x0) * del-list(rx0, m - # x0)) * P))(H0 ∪ H1)) ↔ n0 + # x = m ∧ P(H1), r = null

Rule argument

spec: heap-sep name: sep seq: ⊦ ((P0 * P1))(H) ↔ (∃ H0, H1. H0 ⊥ H1 ∧ H = H0 ∪ H1 ∧ P0(H0) ∧ P1(H1)) subst: [H, P1, P0] → [H0, del-list(rx0, n0), ls(r, x)] flags: <all paths>

Additional information

Used terms: H0, del-list(rx0, n0), ls(r, x)