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Proof step 4 from proof abs-nodes-case-distinction-h (previous step)

x = [], 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 5:
abs-nodes(buf, m, [], [], true) ⊦ buf = mkbuf(0), abs-failed(buf), ∃ buf0, buf1. buf = buf0 + buf1 ∧ abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, [], [], true)

Rule argument

Additional information

Used simplifier rules: abs-nodes-elems-empty-adr-empty-h: ⊦ abs-nodes(buf, m, [], lax, weak) → lax = [];