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Proof step 9 from proof abs-nodes-nodups-h (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
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 ⊦

Proof step 9 from proof abs-nodes-nodups-h (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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