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Proof step 15 from proof abs-nodes-prefix-h (previous step)

lax0.head = @(m0, m1), m0 = m, m1 = # buf2, abs-node(buf2, nd), abs-nodes((buf2 + buf) + buf1, m, x1, lax1, boolvar1), abs-nodes(buf, m + # buf2, x, lax0.tail, boolvar0), ∀ buf3, buf1, m, x1, lax1, boolvar1, x0, lax0, boolvar0. # buf3 < #(buf2 + buf) → abs-nodes(buf3 + buf1, m, x1, lax1, boolvar1) ∧ abs-nodes(buf3, m, x0, lax0, boolvar0) → x0 ⊑ x1 ∧ lax0 ⊑ lax1 ⊦ lax0 = [], nd ' + x ⊑ x1 ∧ lax0 ⊑ lax1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# buf2 = nd.alignedsize + 2 * NODE_HEADER_SIZE, lax0.head = @(m, # buf2), abs-node(buf2, nd), abs-nodes(buf, m + # buf2, x, lax0.tail, boolvar0), abs-nodes((buf2 + buf) + buf1, m, x1, lax1, boolvar1), ∀ buf3, buf1, m, x1, lax1, boolvar1, x0, lax0, boolvar0. # buf3 < #(buf2 + buf) → abs-nodes(buf3 + buf1, m, x1, lax1, boolvar1) ∧ abs-nodes(buf3, m, x0, lax0, boolvar0) → x0 ⊑ x1 ∧ lax0 ⊑ lax1 ⊦ lax0 = [], nd ' + x ⊑ x1 ∧ lax0 ⊑ lax1

Rule argument

Additional information

Used simplifier rules: ls-abs-node-buf-size: ⊦ abs-node(buf, nd) → # buf = nd.alignedsize + 2 * NODE_HEADER_SIZE;