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Proof step 9 from proof abs-nodes-no-padding-rec (previous step)

# buf = (2 * NODE_HEADER_SIZE + a.alignedsize) + x.flashsize, abs-nodes(buf, m, a ' + x, @(m0, m1) ' + lax0, false), ∀ buf0, a, x, m, m0, m1, lax0. # buf0 < # buf → # buf0 = (2 * NODE_HEADER_SIZE + a.alignedsize) + x.flashsize ∧ abs-nodes(buf0, m, a ' + x, @(m0, m1) ' + lax0, false) → (∃ buf2, buf1. buf0 = buf2 + buf1 ∧ m0 = m ∧ m1 = # buf2 ∧ abs-node(buf2, a) ∧ abs-nodes(buf1, m + # buf2, x, lax0, false)) ⊦ ∃ buf0, buf1. buf = buf0 + buf1 ∧ m0 = m ∧ m1 = # buf0 ∧ abs-node(buf0, a) ∧ abs-nodes(buf1, m + # buf0, x, lax0, false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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