Left Termination of the query pattern tree_member_in_2(g, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

tree_member(X, tree(X, X1, X2)).
tree_member(X, tree(X1, Left, X2)) :- tree_member(X, Left).
tree_member(X, tree(X1, X2, Right)) :- tree_member(X, Right).

Queries:

tree_member(g,a).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
tree_member_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → U1_GA(X, X1, Left, X2, tree_member_in_ga(X, Left))
TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → TREE_MEMBER_IN_GA(X, Left)
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → U2_GA(X, X1, X2, Right, tree_member_in_ga(X, Right))
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → TREE_MEMBER_IN_GA(X, Right)

The TRS R consists of the following rules:

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x5)
TREE_MEMBER_IN_GA(x1, x2)  =  TREE_MEMBER_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x5)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → U1_GA(X, X1, Left, X2, tree_member_in_ga(X, Left))
TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → TREE_MEMBER_IN_GA(X, Left)
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → U2_GA(X, X1, X2, Right, tree_member_in_ga(X, Right))
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → TREE_MEMBER_IN_GA(X, Right)

The TRS R consists of the following rules:

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x5)
TREE_MEMBER_IN_GA(x1, x2)  =  TREE_MEMBER_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → TREE_MEMBER_IN_GA(X, Left)
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → TREE_MEMBER_IN_GA(X, Right)

The TRS R consists of the following rules:

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x5)
TREE_MEMBER_IN_GA(x1, x2)  =  TREE_MEMBER_IN_GA(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

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

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

TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → TREE_MEMBER_IN_GA(X, Left)
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → TREE_MEMBER_IN_GA(X, Right)

R is empty.
The argument filtering Pi contains the following mapping:
TREE_MEMBER_IN_GA(x1, x2)  =  TREE_MEMBER_IN_GA(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

TREE_MEMBER_IN_GA(X) → TREE_MEMBER_IN_GA(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

TREE_MEMBER_IN_GA(X) → TREE_MEMBER_IN_GA(X)

The TRS R consists of the following rules:none


s = TREE_MEMBER_IN_GA(X) evaluates to t =TREE_MEMBER_IN_GA(X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from TREE_MEMBER_IN_GA(X) to TREE_MEMBER_IN_GA(X).




We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
tree_member_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga(x1)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga(x1)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x5)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → U1_GA(X, X1, Left, X2, tree_member_in_ga(X, Left))
TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → TREE_MEMBER_IN_GA(X, Left)
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → U2_GA(X, X1, X2, Right, tree_member_in_ga(X, Right))
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → TREE_MEMBER_IN_GA(X, Right)

The TRS R consists of the following rules:

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga(x1)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x5)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x5)
TREE_MEMBER_IN_GA(x1, x2)  =  TREE_MEMBER_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x5)

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

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → U1_GA(X, X1, Left, X2, tree_member_in_ga(X, Left))
TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → TREE_MEMBER_IN_GA(X, Left)
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → U2_GA(X, X1, X2, Right, tree_member_in_ga(X, Right))
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → TREE_MEMBER_IN_GA(X, Right)

The TRS R consists of the following rules:

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga(x1)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x5)
U2_GA(x1, x2, x3, x4, x5)  =  U2_GA(x1, x5)
TREE_MEMBER_IN_GA(x1, x2)  =  TREE_MEMBER_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5)  =  U1_GA(x1, x5)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 2 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → TREE_MEMBER_IN_GA(X, Left)
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → TREE_MEMBER_IN_GA(X, Right)

The TRS R consists of the following rules:

tree_member_in_ga(X, tree(X, X1, X2)) → tree_member_out_ga(X, tree(X, X1, X2))
tree_member_in_ga(X, tree(X1, Left, X2)) → U1_ga(X, X1, Left, X2, tree_member_in_ga(X, Left))
tree_member_in_ga(X, tree(X1, X2, Right)) → U2_ga(X, X1, X2, Right, tree_member_in_ga(X, Right))
U2_ga(X, X1, X2, Right, tree_member_out_ga(X, Right)) → tree_member_out_ga(X, tree(X1, X2, Right))
U1_ga(X, X1, Left, X2, tree_member_out_ga(X, Left)) → tree_member_out_ga(X, tree(X1, Left, X2))

The argument filtering Pi contains the following mapping:
tree_member_in_ga(x1, x2)  =  tree_member_in_ga(x1)
tree_member_out_ga(x1, x2)  =  tree_member_out_ga(x1)
U1_ga(x1, x2, x3, x4, x5)  =  U1_ga(x1, x5)
U2_ga(x1, x2, x3, x4, x5)  =  U2_ga(x1, x5)
TREE_MEMBER_IN_GA(x1, x2)  =  TREE_MEMBER_IN_GA(x1)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

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

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

TREE_MEMBER_IN_GA(X, tree(X1, Left, X2)) → TREE_MEMBER_IN_GA(X, Left)
TREE_MEMBER_IN_GA(X, tree(X1, X2, Right)) → TREE_MEMBER_IN_GA(X, Right)

R is empty.
The argument filtering Pi contains the following mapping:
TREE_MEMBER_IN_GA(x1, x2)  =  TREE_MEMBER_IN_GA(x1)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ NonTerminationProof

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

TREE_MEMBER_IN_GA(X) → TREE_MEMBER_IN_GA(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

TREE_MEMBER_IN_GA(X) → TREE_MEMBER_IN_GA(X)

The TRS R consists of the following rules:none


s = TREE_MEMBER_IN_GA(X) evaluates to t =TREE_MEMBER_IN_GA(X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from TREE_MEMBER_IN_GA(X) to TREE_MEMBER_IN_GA(X).