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Proof step 22 from proof abs-split-by-address-h (previous step)

abs-padding(buf1), abs-padding(buf2) ∧ abs-nodes(buf, (m + # buf1) + # buf2, nd ' + x2, @(m0, m1) ' + lax3, boolvar0) ∨ m0 = m + # buf1 ∧ m1 = # buf2 ∧ abs-node(buf2, nd) ∧ abs-nodes(buf, (m + # buf1) + # buf2, x2, lax3, boolvar0), ∀ buf3, m, x, lax0, ladr, lax1, boolvar0. # buf3 < #(buf1 + buf2 + buf) → abs-nodes(buf3, m, x, lax0 + ladr ' + lax1, boolvar0) → (∃ buf5, buf1, buf4, x0, a, x1. x = x0 + a ' + x1 ∧ # x0 = # lax0 ∧ buf3 = buf5 + buf1 + buf4 ∧ ladr = @(m + # buf5, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf1 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf1, a)) ⊦ ∃ buf5, buf4, buf3, a, x0. nd ' + x2 = a ' + x0 ∧ buf1 + buf2 + buf = buf5 + buf4 + buf3 ∧ @(m0, m1) = @(m + # buf5, 2 * NODE_HEADER_SIZE + a.alignedsize) ∧ # buf4 = 2 * NODE_HEADER_SIZE + a.alignedsize ∧ abs-node(buf4, a)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

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