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Proof step 17 from proof del-lists-unions (previous step)

(((ls(r, x0) * del-list(rx0, m - # x0)) * P))((H2 ∪ H) ∪ H1), del-list(rx0, n0)(H), ls(r, x)(H2), # x0 ≤ m, 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, n0 + # x = m ∧ P(H1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
m = n1 + # x0, (((ls(r, x0) * del-list(rx0, n1)) * P))((H2 ∪ H) ∪ H1), del-list(rx0, n0)(H), ls(r, x)(H2), # x0 ≤ m, H1 ⊥ H, H1 ⊥ H2, H2 ⊥ 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, n0 + # x = m ∧ P(H1)

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [n, m, n0] → [# x0, m, n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # x0, m, n1