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Proof step 6 from proof abs-nodes-no-padding-last (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Rule argument

var: x

Additional information

Used simplifier rules: list freely generated by [], +;;