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Proof step 35 from proof del-lists-unions (previous step)

P(H1), del-list(rx0, n0)(H), ls(r, x)(H2), H1 ⊥ H2, H2 ⊥ H, 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 36:
⊦ H0 ⊥ H1 ∧ ls(r, x)(H0) → (((ls(r, y) * P))(H0 ∪ H1) ↔ x = y ∧ P(H1))
Proof step 37:
P(H1), del-list(rx0, n0)(H), ls(r, x)(H2), H1 ⊥ H, H2 ⊥ H, 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), H2 ⊥ H ∪ H1 ∧ ls(r, x)(H2)
Proof step 40:
P(H1), del-list(rx0, n0)(H), ls(r, x)(H2), H1 ⊥ H, H2 ⊥ H, H1 ⊥ H2, H2 ⊥ H ∪ H1, ∀ 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, x = x ∧ ((P * del-list(rx0, n0)))(H ∪ H1)

Rule argument

spec: linked-list name: ls-union seq: ⊦ H0 ⊥ H1 ∧ ls(r, x)(H0) → (((ls(r, y) * P))(H0 ∪ H1) ↔ x = y ∧ P(H1)) subst: [x, H1, H0, P, y, r] → [x, H ∪ H1, H2, P * del-list(rx0, n0), x, r] flags: <all paths>

Additional information

Used simplifier rules: sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep Used terms: x, H ∪ H1, H2, P * del-list(rx0, n0), x, r