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Proof step 6 from proof abs-nodes-split (previous step)

abs-nodes(buf1, # buf0, x1, lax1, weak), abs-nodes(buf0, 0, x0, lax0, false), abs-nodes(buf0 + buf1, 0, x0 + x1, lax0 + lax1, weak) ⊦ ∃ buf2, buf3, lax2, lax3. lax0 + lax1 = lax2 + lax3 ∧ buf0 + buf1 = buf2 + buf3 ∧ abs-nodes(buf2, 0, x0, lax2, false) ∧ abs-nodes(buf3, # buf2, x1, lax3, weak)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf0, buf1, lax0, lax1

Additional information

Used simplifier rules: abs-nodes-unique-strong-h: ⊦ abs-nodes(buf, m, x0, lax0, false) → (abs-nodes(buf, m, x1, lax1, boolvar1) ↔ x0 = x1 ∧ lax0 = lax1); Used terms: buf0, buf1, lax0, lax1