Left Termination of the query pattern f_in_3(a, a, g) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

f(0, 1, X) :- f(X, X, X).

Queries:

f(a,a,g).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x2)
f_out(x1, x2, x3)  =  f_out(x1, x2)

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:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x2)
f_out(x1, x2, x3)  =  f_out(x1, x2)


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

F_IN(0, 1, X) → U11(X, f_in(X, X, X))
F_IN(0, 1, X) → F_IN(X, X, X)

The TRS R consists of the following rules:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x2)
f_out(x1, x2, x3)  =  f_out(x1, x2)
F_IN(x1, x2, x3)  =  F_IN(x3)
U11(x1, x2)  =  U11(x2)

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:

F_IN(0, 1, X) → U11(X, f_in(X, X, X))
F_IN(0, 1, X) → F_IN(X, X, X)

The TRS R consists of the following rules:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x2)
f_out(x1, x2, x3)  =  f_out(x1, x2)
F_IN(x1, x2, x3)  =  F_IN(x3)
U11(x1, x2)  =  U11(x2)

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

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

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

F_IN(0, 1, X) → F_IN(X, X, X)

The TRS R consists of the following rules:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x2)
f_out(x1, x2, x3)  =  f_out(x1, x2)
F_IN(x1, x2, x3)  =  F_IN(x3)

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:

F_IN(0, 1, X) → F_IN(X, X, X)

R is empty.
The argument filtering Pi contains the following mapping:
0  =  0
1  =  1
F_IN(x1, x2, x3)  =  F_IN(x3)

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:

F_IN(X) → F_IN(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:

F_IN(X) → F_IN(X)

The TRS R consists of the following rules:none


s = F_IN(X) evaluates to t =F_IN(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 F_IN(X) to F_IN(X).




We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x1, x2)
f_out(x1, x2, x3)  =  f_out(x1, x2, x3)

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:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x1, x2)
f_out(x1, x2, x3)  =  f_out(x1, x2, x3)


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

F_IN(0, 1, X) → U11(X, f_in(X, X, X))
F_IN(0, 1, X) → F_IN(X, X, X)

The TRS R consists of the following rules:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x1, x2)
f_out(x1, x2, x3)  =  f_out(x1, x2, x3)
F_IN(x1, x2, x3)  =  F_IN(x3)
U11(x1, x2)  =  U11(x1, x2)

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:

F_IN(0, 1, X) → U11(X, f_in(X, X, X))
F_IN(0, 1, X) → F_IN(X, X, X)

The TRS R consists of the following rules:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x1, x2)
f_out(x1, x2, x3)  =  f_out(x1, x2, x3)
F_IN(x1, x2, x3)  =  F_IN(x3)
U11(x1, x2)  =  U11(x1, x2)

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

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

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

F_IN(0, 1, X) → F_IN(X, X, X)

The TRS R consists of the following rules:

f_in(0, 1, X) → U1(X, f_in(X, X, X))
U1(X, f_out(X, X, X)) → f_out(0, 1, X)

The argument filtering Pi contains the following mapping:
f_in(x1, x2, x3)  =  f_in(x3)
0  =  0
1  =  1
U1(x1, x2)  =  U1(x1, x2)
f_out(x1, x2, x3)  =  f_out(x1, x2, x3)
F_IN(x1, x2, x3)  =  F_IN(x3)

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:

F_IN(0, 1, X) → F_IN(X, X, X)

R is empty.
The argument filtering Pi contains the following mapping:
0  =  0
1  =  1
F_IN(x1, x2, x3)  =  F_IN(x3)

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:

F_IN(X) → F_IN(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:

F_IN(X) → F_IN(X)

The TRS R consists of the following rules:none


s = F_IN(X) evaluates to t =F_IN(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 F_IN(X) to F_IN(X).