Left Termination of the query pattern thief_in_1(g) w.r.t. the given Prolog program could not be shown:

(0) Obligation:

Clauses:

thief(john).
thief(X) :- ','(thief(X), !).

Queries:

thief(g).

(1) CutEliminatorProof (SOUND transformation)

Eliminated all cuts by simply ignoring them[PROLOG].

(2) Obligation:

Clauses:

thief(john).
thief(X) :- thief(X).

Queries:

thief(g).

(3) PrologToPiTRSProof (SOUND transformation)

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

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

The argument filtering Pi contains the following mapping:
thief_in_g(x1)  =  thief_in_g(x1)
john  =  john
thief_out_g(x1)  =  thief_out_g
U1_g(x1, x2)  =  U1_g(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

The argument filtering Pi contains the following mapping:
thief_in_g(x1)  =  thief_in_g(x1)
john  =  john
thief_out_g(x1)  =  thief_out_g
U1_g(x1, x2)  =  U1_g(x2)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

THIEF_IN_G(X) → U1_G(X, thief_in_g(X))
THIEF_IN_G(X) → THIEF_IN_G(X)

The TRS R consists of the following rules:

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

The argument filtering Pi contains the following mapping:
thief_in_g(x1)  =  thief_in_g(x1)
john  =  john
thief_out_g(x1)  =  thief_out_g
U1_g(x1, x2)  =  U1_g(x2)
THIEF_IN_G(x1)  =  THIEF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(6) Obligation:

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

THIEF_IN_G(X) → U1_G(X, thief_in_g(X))
THIEF_IN_G(X) → THIEF_IN_G(X)

The TRS R consists of the following rules:

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

The argument filtering Pi contains the following mapping:
thief_in_g(x1)  =  thief_in_g(x1)
john  =  john
thief_out_g(x1)  =  thief_out_g
U1_g(x1, x2)  =  U1_g(x2)
THIEF_IN_G(x1)  =  THIEF_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(8) Obligation:

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

THIEF_IN_G(X) → THIEF_IN_G(X)

The TRS R consists of the following rules:

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

The argument filtering Pi contains the following mapping:
thief_in_g(x1)  =  thief_in_g(x1)
john  =  john
thief_out_g(x1)  =  thief_out_g
U1_g(x1, x2)  =  U1_g(x2)
THIEF_IN_G(x1)  =  THIEF_IN_G(x1)

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

(9) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(10) Obligation:

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

THIEF_IN_G(X) → THIEF_IN_G(X)

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

(11) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(12) Obligation:

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

THIEF_IN_G(X) → THIEF_IN_G(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(13) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

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

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

• Semiunifier: [ ]
• Matcher: [ ]

---

Rewriting sequence

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

(14) FALSE

(15) PrologToPiTRSProof (SOUND transformation)

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

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

Pi is empty.

(17) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

THIEF_IN_G(X) → U1_G(X, thief_in_g(X))
THIEF_IN_G(X) → THIEF_IN_G(X)

The TRS R consists of the following rules:

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

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

(18) Obligation:

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

THIEF_IN_G(X) → U1_G(X, thief_in_g(X))
THIEF_IN_G(X) → THIEF_IN_G(X)

The TRS R consists of the following rules:

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(20) Obligation:

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

THIEF_IN_G(X) → THIEF_IN_G(X)

The TRS R consists of the following rules:

thief_in_g(john) → thief_out_g(john)
thief_in_g(X) → U1_g(X, thief_in_g(X))
U1_g(X, thief_out_g(X)) → thief_out_g(X)

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

(21) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(22) Obligation:

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

THIEF_IN_G(X) → THIEF_IN_G(X)

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

(23) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(24) Obligation:

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

THIEF_IN_G(X) → THIEF_IN_G(X)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(25) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

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

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

• Matcher: [ ]
• Semiunifier: [ ]

---

Rewriting sequence

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

(26) FALSE