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Proof step 18 from proof abs-nodes-padding-prefix (previous step)

abs-nodes(buf1, m + # buf0, nd ' + x0, @(m0, m1) ' + lax0, weak), abs-padding(buf0) ⊦ ∃ buf, buf2. buf0 + buf1 = buf + buf2 ∧ ( buf0 = buf ∧ abs-nodes(buf2, m + # buf, nd ' + x0, @(m0, m1) ' + lax0, weak) ∨ m0 = m ∧ m1 = # buf ∧ abs-node(buf, nd) ∧ abs-nodes(buf2, m + # buf, x0, lax0, weak))

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf0, buf1

Additional information

Used terms: buf0, buf1