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Proof step 13 from proof abs-nodes-split-h (previous step)

lax.head = @(m0, m1), abs-padding(buf1) ∧ abs-nodes(buf0, m + # buf1, nd ' + x + x1, lax, weak) ∨ lax ≠ [] ∧ m0 = m ∧ m1 = # buf1 ∧ abs-node(buf1, nd) ∧ abs-nodes(buf0, m + # buf1, x + x1, lax.tail, weak), ∀ buf2, m, x0, x1, lax, weak. # buf2 < #(buf1 + buf0) → abs-nodes(buf2, m, x0 + x1, lax, weak) → (∃ buf3, buf1, lax0, lax1. lax = lax0 + lax1 ∧ buf2 = buf3 + buf1 ∧ abs-nodes(buf3, m, x0, lax0, false) ∧ abs-nodes(buf1, m + # buf3, x1, lax1, weak)) ⊦ ∃ buf3, buf2, lax0, lax1. lax = lax0 + lax1 ∧ buf1 + buf0 = buf3 + buf2 ∧ abs-nodes(buf3, m, nd ' + x, lax0, false) ∧ abs-nodes(buf2, m + # buf3, x1, lax1, weak)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 2

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