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Proof step 7 from proof abs-nodes-case-distinction-h (previous step)

x = nd + x0, abs-nodes(buf, m, x, lax, true) ⊦ ∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, x, lax, true) ∨ x ≠ [] ∧ lax ≠ [] ∧ lax.head = @(m, 2 * NODE_HEADER_SIZE + x.head.alignedsize) ∧ abs-node(buf0, x.head) ∧ abs-nodes(buf1, m + # buf0, x.tail, lax.tail, true)), x = [] ∧ buf = mkbuf(0), x = [] ∧ abs-failed(buf)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
abs-nodes(buf, m, nd ' + x0, lax, 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))

Rule argument

Additional information

Used simplifier rules: abs-nodes-first-adr-explicit: ⊦ abs-nodes(buf, m, a ' + x, lax, weak) → (lax.head = ladr ↔ (∃ lax0. lax = ladr + lax0));% ⊦ (a ' + x).tail = x; (s)% ⊦ ladr + lax = ladr ' + lax; (s)% ⊦ (a ' + x).head = a; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)