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Proof step 10 from proof abs-nodes-case-distinction-h (previous step)

∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, nd ' + x0, lax, true) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf0 ∧ abs-node(buf0, nd) ∧ abs-nodes(buf1, m + # buf0, x0, lax.tail, true)) ⊦ ∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, nd ' + x0, lax, true) ∨ lax ≠ [] ∧ (∃ lax0. lax = @(m, 2 * NODE_HEADER_SIZE + nd.alignedsize) ' + lax0) ∧ abs-node(buf0, nd) ∧ abs-nodes(buf1, m + # buf0, x0, lax.tail, true))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x0, lax, true) ∨ lax ≠ [] ∧ lax.head.off = m ∧ lax.head.size = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, lax.head.off + # buf1, x0, lax.tail, true) ⊦ ∃ buf3, buf2. buf1 + buf0 = buf3 + buf2 ∧ ( abs-padding(buf3) ∧ abs-nodes(buf2, m + # buf3, nd ' + x0, lax, true) ∨ lax ≠ [] ∧ (∃ lax0. lax = @(m, 2 * NODE_HEADER_SIZE + nd.alignedsize) ' + lax0) ∧ abs-node(buf3, nd) ∧ abs-nodes(buf2, m + # buf3, x0, lax.tail, true))

Proof step 10 from proof abs-nodes-case-distinction-h (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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