Left Termination of the query pattern member_in_2(g, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

member(X, .(X, X1)).
member(X, .(X1, Xs)) :- member(X, Xs).

Queries:

member(g,a).

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:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
member_out(x1, x2)  =  member_out

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:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
member_out(x1, x2)  =  member_out


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

MEMBER_IN(X, .(X1, Xs)) → U11(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
member_out(x1, x2)  =  member_out
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x4)

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:

MEMBER_IN(X, .(X1, Xs)) → U11(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
member_out(x1, x2)  =  member_out
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x4)

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:

MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x4)
member_out(x1, x2)  =  member_out
MEMBER_IN(x1, x2)  =  MEMBER_IN(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:

MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

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

MEMBER_IN(X) → MEMBER_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:

MEMBER_IN(X) → MEMBER_IN(X)

The TRS R consists of the following rules:none


s = MEMBER_IN(X) evaluates to t =MEMBER_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 MEMBER_IN(X) to MEMBER_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:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
member_out(x1, x2)  =  member_out(x1)

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:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
member_out(x1, x2)  =  member_out(x1)


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

MEMBER_IN(X, .(X1, Xs)) → U11(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
member_out(x1, x2)  =  member_out(x1)
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x1, x4)

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:

MEMBER_IN(X, .(X1, Xs)) → U11(X, X1, Xs, member_in(X, Xs))
MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
member_out(x1, x2)  =  member_out(x1)
MEMBER_IN(x1, x2)  =  MEMBER_IN(x1)
U11(x1, x2, x3, x4)  =  U11(x1, x4)

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:

MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

The TRS R consists of the following rules:

member_in(X, .(X1, Xs)) → U1(X, X1, Xs, member_in(X, Xs))
member_in(X, .(X, X1)) → member_out(X, .(X, X1))
U1(X, X1, Xs, member_out(X, Xs)) → member_out(X, .(X1, Xs))

The argument filtering Pi contains the following mapping:
member_in(x1, x2)  =  member_in(x1)
.(x1, x2)  =  .(x1, x2)
U1(x1, x2, x3, x4)  =  U1(x1, x4)
member_out(x1, x2)  =  member_out(x1)
MEMBER_IN(x1, x2)  =  MEMBER_IN(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:

MEMBER_IN(X, .(X1, Xs)) → MEMBER_IN(X, Xs)

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

MEMBER_IN(X) → MEMBER_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:

MEMBER_IN(X) → MEMBER_IN(X)

The TRS R consists of the following rules:none


s = MEMBER_IN(X) evaluates to t =MEMBER_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 MEMBER_IN(X) to MEMBER_IN(X).