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Proof step 11 from proof abs-nodes-nodups-h (previous step)

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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
ladr ∈ lax0 ∨ dups(lax0), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, ladr ' + lax0, boolvar) ∨ 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 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ dups(ladr ' + lax) ↔ ladr ∈ lax ∨ dups(lax); (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ ladr ' + lax ≠ []; (s)% ⊦ ladr + lax = ladr ' + lax; (s)