Left Termination of the query pattern tree_member(f,b) w.r.t. the given Prolog program could successfully be proven:



↳ PROLOG
  ↳ PrologToPiTRSProof

treemember2(X, tree3(X, underscore, underscore1)).
treemember2(X, tree3(underscore2, Left, underscore3)) :- treemember2(X, Left).
treemember2(X, tree3(underscore4, underscore5, Right)) :- treemember2(X, Right).


With regard to the inferred argument filtering the predicates were used in the following modes:
tree_member2: (f,b)
Transforming PROLOG into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:


tree_member_2_in_ag2(X, tree_33(X, underscore, underscore1)) -> tree_member_2_out_ag2(X, tree_33(X, underscore, underscore1))
tree_member_2_in_ag2(X, tree_33(underscore2, Left, underscore3)) -> if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_in_ag2(X, Left))
tree_member_2_in_ag2(X, tree_33(underscore4, underscore5, Right)) -> if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_in_ag2(X, Right))
if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_out_ag2(X, Right)) -> tree_member_2_out_ag2(X, tree_33(underscore4, underscore5, Right))
if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_out_ag2(X, Left)) -> tree_member_2_out_ag2(X, tree_33(underscore2, Left, underscore3))

The argument filtering Pi contains the following mapping:
tree_member_2_in_ag2(x1, x2)  =  tree_member_2_in_ag1(x2)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
tree_member_2_out_ag2(x1, x2)  =  tree_member_2_out_ag1(x1)
if_tree_member_2_in_1_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_1_ag1(x5)
if_tree_member_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_2_ag1(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of PROLOG



↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

Pi-finite rewrite system:
The TRS R consists of the following rules:

tree_member_2_in_ag2(X, tree_33(X, underscore, underscore1)) -> tree_member_2_out_ag2(X, tree_33(X, underscore, underscore1))
tree_member_2_in_ag2(X, tree_33(underscore2, Left, underscore3)) -> if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_in_ag2(X, Left))
tree_member_2_in_ag2(X, tree_33(underscore4, underscore5, Right)) -> if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_in_ag2(X, Right))
if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_out_ag2(X, Right)) -> tree_member_2_out_ag2(X, tree_33(underscore4, underscore5, Right))
if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_out_ag2(X, Left)) -> tree_member_2_out_ag2(X, tree_33(underscore2, Left, underscore3))

The argument filtering Pi contains the following mapping:
tree_member_2_in_ag2(x1, x2)  =  tree_member_2_in_ag1(x2)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
tree_member_2_out_ag2(x1, x2)  =  tree_member_2_out_ag1(x1)
if_tree_member_2_in_1_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_1_ag1(x5)
if_tree_member_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_2_ag1(x5)


Pi DP problem:
The TRS P consists of the following rules:

TREE_MEMBER_2_IN_AG2(X, tree_33(underscore2, Left, underscore3)) -> IF_TREE_MEMBER_2_IN_1_AG5(X, underscore2, Left, underscore3, tree_member_2_in_ag2(X, Left))
TREE_MEMBER_2_IN_AG2(X, tree_33(underscore2, Left, underscore3)) -> TREE_MEMBER_2_IN_AG2(X, Left)
TREE_MEMBER_2_IN_AG2(X, tree_33(underscore4, underscore5, Right)) -> IF_TREE_MEMBER_2_IN_2_AG5(X, underscore4, underscore5, Right, tree_member_2_in_ag2(X, Right))
TREE_MEMBER_2_IN_AG2(X, tree_33(underscore4, underscore5, Right)) -> TREE_MEMBER_2_IN_AG2(X, Right)

The TRS R consists of the following rules:

tree_member_2_in_ag2(X, tree_33(X, underscore, underscore1)) -> tree_member_2_out_ag2(X, tree_33(X, underscore, underscore1))
tree_member_2_in_ag2(X, tree_33(underscore2, Left, underscore3)) -> if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_in_ag2(X, Left))
tree_member_2_in_ag2(X, tree_33(underscore4, underscore5, Right)) -> if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_in_ag2(X, Right))
if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_out_ag2(X, Right)) -> tree_member_2_out_ag2(X, tree_33(underscore4, underscore5, Right))
if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_out_ag2(X, Left)) -> tree_member_2_out_ag2(X, tree_33(underscore2, Left, underscore3))

The argument filtering Pi contains the following mapping:
tree_member_2_in_ag2(x1, x2)  =  tree_member_2_in_ag1(x2)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
tree_member_2_out_ag2(x1, x2)  =  tree_member_2_out_ag1(x1)
if_tree_member_2_in_1_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_1_ag1(x5)
if_tree_member_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_2_ag1(x5)
TREE_MEMBER_2_IN_AG2(x1, x2)  =  TREE_MEMBER_2_IN_AG1(x2)
IF_TREE_MEMBER_2_IN_2_AG5(x1, x2, x3, x4, x5)  =  IF_TREE_MEMBER_2_IN_2_AG1(x5)
IF_TREE_MEMBER_2_IN_1_AG5(x1, x2, x3, x4, x5)  =  IF_TREE_MEMBER_2_IN_1_AG1(x5)

We have to consider all (P,R,Pi)-chains

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

Pi DP problem:
The TRS P consists of the following rules:

TREE_MEMBER_2_IN_AG2(X, tree_33(underscore2, Left, underscore3)) -> IF_TREE_MEMBER_2_IN_1_AG5(X, underscore2, Left, underscore3, tree_member_2_in_ag2(X, Left))
TREE_MEMBER_2_IN_AG2(X, tree_33(underscore2, Left, underscore3)) -> TREE_MEMBER_2_IN_AG2(X, Left)
TREE_MEMBER_2_IN_AG2(X, tree_33(underscore4, underscore5, Right)) -> IF_TREE_MEMBER_2_IN_2_AG5(X, underscore4, underscore5, Right, tree_member_2_in_ag2(X, Right))
TREE_MEMBER_2_IN_AG2(X, tree_33(underscore4, underscore5, Right)) -> TREE_MEMBER_2_IN_AG2(X, Right)

The TRS R consists of the following rules:

tree_member_2_in_ag2(X, tree_33(X, underscore, underscore1)) -> tree_member_2_out_ag2(X, tree_33(X, underscore, underscore1))
tree_member_2_in_ag2(X, tree_33(underscore2, Left, underscore3)) -> if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_in_ag2(X, Left))
tree_member_2_in_ag2(X, tree_33(underscore4, underscore5, Right)) -> if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_in_ag2(X, Right))
if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_out_ag2(X, Right)) -> tree_member_2_out_ag2(X, tree_33(underscore4, underscore5, Right))
if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_out_ag2(X, Left)) -> tree_member_2_out_ag2(X, tree_33(underscore2, Left, underscore3))

The argument filtering Pi contains the following mapping:
tree_member_2_in_ag2(x1, x2)  =  tree_member_2_in_ag1(x2)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
tree_member_2_out_ag2(x1, x2)  =  tree_member_2_out_ag1(x1)
if_tree_member_2_in_1_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_1_ag1(x5)
if_tree_member_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_2_ag1(x5)
TREE_MEMBER_2_IN_AG2(x1, x2)  =  TREE_MEMBER_2_IN_AG1(x2)
IF_TREE_MEMBER_2_IN_2_AG5(x1, x2, x3, x4, x5)  =  IF_TREE_MEMBER_2_IN_2_AG1(x5)
IF_TREE_MEMBER_2_IN_1_AG5(x1, x2, x3, x4, x5)  =  IF_TREE_MEMBER_2_IN_1_AG1(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

Pi DP problem:
The TRS P consists of the following rules:

TREE_MEMBER_2_IN_AG2(X, tree_33(underscore4, underscore5, Right)) -> TREE_MEMBER_2_IN_AG2(X, Right)
TREE_MEMBER_2_IN_AG2(X, tree_33(underscore2, Left, underscore3)) -> TREE_MEMBER_2_IN_AG2(X, Left)

The TRS R consists of the following rules:

tree_member_2_in_ag2(X, tree_33(X, underscore, underscore1)) -> tree_member_2_out_ag2(X, tree_33(X, underscore, underscore1))
tree_member_2_in_ag2(X, tree_33(underscore2, Left, underscore3)) -> if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_in_ag2(X, Left))
tree_member_2_in_ag2(X, tree_33(underscore4, underscore5, Right)) -> if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_in_ag2(X, Right))
if_tree_member_2_in_2_ag5(X, underscore4, underscore5, Right, tree_member_2_out_ag2(X, Right)) -> tree_member_2_out_ag2(X, tree_33(underscore4, underscore5, Right))
if_tree_member_2_in_1_ag5(X, underscore2, Left, underscore3, tree_member_2_out_ag2(X, Left)) -> tree_member_2_out_ag2(X, tree_33(underscore2, Left, underscore3))

The argument filtering Pi contains the following mapping:
tree_member_2_in_ag2(x1, x2)  =  tree_member_2_in_ag1(x2)
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
tree_member_2_out_ag2(x1, x2)  =  tree_member_2_out_ag1(x1)
if_tree_member_2_in_1_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_1_ag1(x5)
if_tree_member_2_in_2_ag5(x1, x2, x3, x4, x5)  =  if_tree_member_2_in_2_ag1(x5)
TREE_MEMBER_2_IN_AG2(x1, x2)  =  TREE_MEMBER_2_IN_AG1(x2)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting we can delete all non-usable rules from R.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

Pi DP problem:
The TRS P consists of the following rules:

TREE_MEMBER_2_IN_AG2(X, tree_33(underscore4, underscore5, Right)) -> TREE_MEMBER_2_IN_AG2(X, Right)
TREE_MEMBER_2_IN_AG2(X, tree_33(underscore2, Left, underscore3)) -> TREE_MEMBER_2_IN_AG2(X, Left)

R is empty.
The argument filtering Pi contains the following mapping:
tree_33(x1, x2, x3)  =  tree_33(x1, x2, x3)
TREE_MEMBER_2_IN_AG2(x1, x2)  =  TREE_MEMBER_2_IN_AG1(x2)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem into ordinary QDP problem by application of Pi.

↳ PROLOG
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ QDPSizeChangeProof

Q DP problem:
The TRS P consists of the following rules:

TREE_MEMBER_2_IN_AG1(tree_33(underscore4, underscore5, Right)) -> TREE_MEMBER_2_IN_AG1(Right)
TREE_MEMBER_2_IN_AG1(tree_33(underscore2, Left, underscore3)) -> TREE_MEMBER_2_IN_AG1(Left)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
The head symbols of this DP problem are {TREE_MEMBER_2_IN_AG1}.
By using the subterm criterion together with the size-change analysis we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: