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Proof step 8 from proof abs-split-by-address-h (previous step)

∃ buf0, buf1. buf = buf0 + buf1 ∧ ( abs-padding(buf0) ∧ abs-nodes(buf1, m + # buf0, nd ' + x2, lax0 + ladr ' + lax1, boolvar0) ∨ lax0 + ladr ' + lax1 ≠ [] ∧ (lax0 + ladr ' + lax1).head.off = m ∧ (lax0 + ladr ' + lax1).head.size = # buf0 ∧ abs-node(buf0, nd) ∧ abs-nodes(buf1, m + # buf0, x2, (lax0 + ladr ' + lax1).tail, boolvar0)), ∀ buf0, m, x, lax0, ladr, lax1, boolvar0. # buf0 < # buf → abs-nodes(buf0, m, x, lax0 + ladr ' + lax1, boolvar0) → (∃ buf3, buf1, buf2, x0, a, x1. x = x0 + a ' + x1 ∧ # x0 = # lax0 ∧ buf0 = buf3 + buf1 + buf2 ∧ ladr = @(m + # buf3, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf1 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf1, a)) ⊦ ∃ buf0, buf1, buf2, x, a, x0. nd ' + x2 = x + a ' + x0 ∧ # x = # lax0 ∧ buf = buf0 + buf1 + buf2 ∧ ladr = @(m + # buf0, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf1 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf1, a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x2, lax0 + ladr ' + lax1, boolvar0) ∨ (lax0 + ladr ').head.off = m ∧ (lax0 + ladr ').head.size = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, (lax0 + ladr ').head.off + # buf1, x2, (lax0 + ladr ' + lax1).tail, boolvar0), ∀ buf2, m, x, lax0, ladr, lax1, boolvar0. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x, lax0 + ladr ' + lax1, boolvar0) → (∃ buf3, buf1, buf4, x0, a, x1. x = x0 + a ' + x1 ∧ # x0 = # lax0 ∧ buf2 = buf3 + buf1 + buf4 ∧ ladr = @(m + # buf3, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf1 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf1, a)) ⊦ ∃ buf4, buf3, buf2, x, a, x0. nd ' + x2 = x + a ' + x0 ∧ # x = # lax0 ∧ buf1 + buf0 = buf4 + buf3 + buf2 ∧ ladr = @(m + # buf4, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf3 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf3, a)

Rule argument

Additional information

Used simplifier rules: ⊦ (lax + ladr ' + lax0).head = (lax + ladr ').head; (s)%a: ⊦ (lax + lax0) + lax1 = lax + lax0 + lax1; from specification list[laddress-list]% ⊦ lax + ladr ' + lax0 ≠ []; (s)