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Proof step 10 from proof abs-nodes-nodups-h (previous step)

dups(lax), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, lax, boolvar) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, lax.head.off + # buf1, x0, lax.tail, boolvar), ∀ buf2, m, x, lax, boolvar. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, boolvar) ∧ dups(lax) → false ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
lax = ladr + lax0, dups(lax), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, lax, boolvar) ∨ lax ≠ [] ∧ ladr.off = m ∧ ladr.size = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, ladr.off + # buf1, x0, lax0, boolvar), ∀ buf2, m, x, lax, boolvar. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax, boolvar) ∧ dups(lax) → false ⊦ lax = []

Rule argument

spec: list-data, inst: laddress-list name: elim-list-c seq: lax ≠ [] ⊦ ladr = lax.head ∧ lax0 = lax.tail ↔ lax = ladr + lax0 subst: [lax, ladr, lax0] → [lax, ladr, lax0]

Additional information

Used simplifier rules: ⊦ ¬ dups([]); (s)%lax ≠ [] ⊦ ladr = lax.head ∧ lax0 = lax.tail ↔ lax = ladr + lax0; (elim) Used terms: lax, ladr, lax0