Tree Clause Space be characterized cleanly in terms of the ank of the tree underlying the proof. Recall that the rank is defined inductively: leaves have rank 0, and any non-leaf node with subtrees of rank ri and r2 has rank max(r1,r2} if r1r2, and otherwise has rank rit 1 (= r2 + 1). Show that if a treelike resolution refutation has a graph that is of rank k, then the treelike refutation can be carried out (in an appropriate order) using clause space k+1. Tree Clause Space be characterized cleanly in terms of the ank of the tree underlying the proof. Recall that the rank is defined inductively: leaves have rank 0, and any non-leaf node with subtrees of rank ri and r2 has rank max(r1,r2} if r1r2, and otherwise has rank rit 1 (= r2 + 1). Show that if a treelike resolution refutation has a graph that is of rank k, then the treelike refutation can be carried out (in an appropriate order) using clause space k+1