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

lax0.head = @(m0, m1), abs-nodes((buf2 + buf) + buf1, m, x1, lax1, boolvar1), abs-padding(buf2) ∧ abs-nodes(buf, m + # buf2, nd ' + x, lax0, boolvar0) ∨ lax0 ≠ [] ∧ m0 = m ∧ m1 = # buf2 ∧ abs-node(buf2, nd) ∧ abs-nodes(buf, m0 + # buf2, x, lax0.tail, boolvar0), ∀ buf3, buf1, m, x1, lax1, boolvar1, x0, lax0, boolvar0. # buf3 < #(buf2 + buf) → abs-nodes(buf3 + buf1, m, x1, lax1, boolvar1) ∧ abs-nodes(buf3, m, x0, lax0, boolvar0) → x0 ⊑ x1 ∧ lax0 ⊑ lax1 ⊦ nd ' + x ⊑ x1 ∧ lax0 ⊑ lax1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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