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Proof step 21 from proof abs-nodes-one-h (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: buf1, buf0

Additional information

Used terms: buf1, buf0