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Proof step 14 from proof abs-nodes-no-padding-rec (previous step)

# buf1 + # buf0 = (2 * NODE_HEADER_SIZE + a.alignedsize) + x.flashsize, abs-nodes(buf0, m + # buf1, a ' + x, @(m0, m1) ' + lax0, false), abs-padding(buf1), ∀ buf2, a, x, m, m0, m1, lax0. # buf2 < #(buf1 + buf0) → # buf2 = (2 * NODE_HEADER_SIZE + a.alignedsize) + x.flashsize ∧ abs-nodes(buf2, m, a ' + x, @(m0, m1) ' + lax0, false) → (∃ buf3, buf1. buf2 = buf3 + buf1 ∧ m0 = m ∧ m1 = # buf3 ∧ abs-node(buf3, a) ∧ abs-nodes(buf1, m + # buf3, x, lax0, false)) ⊦ ∃ buf3, buf2. buf1 + buf0 = buf3 + buf2 ∧ m0 = m ∧ m1 = # buf3 ∧ abs-node(buf3, a) ∧ abs-nodes(buf2, m + # buf3, x, lax0, false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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