(0) Obligation:

Clauses:

p(X) :- q(X).
p(X) :- p(X).
q(X) :- !.

Queries:

p(a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

p1(T4).
p1(T9).
p1(T11) :- p1(T11).

Queries:

p1(a).

(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:
p1_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(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:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(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:

P1_IN_A(T11) → U1_A(T11, p1_in_a(T11))
P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)

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

(6) Obligation:

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

P1_IN_A(T11) → U1_A(T11, p1_in_a(T11))
P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(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:

P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A

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:

P1_IN_A(T11) → P1_IN_A(T11)

R is empty.
The argument filtering Pi contains the following mapping:
P1_IN_A(x1)  =  P1_IN_A

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

(11) PiDPToQDPProof (SOUND 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:

P1_IN_AP1_IN_A

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 = P1_IN_A evaluates to t =P1_IN_A

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 P1_IN_A to P1_IN_A.



(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:
p1_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)

(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:

P1_IN_A(T11) → U1_A(T11, p1_in_a(T11))
P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)

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

(18) Obligation:

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

P1_IN_A(T11) → U1_A(T11, p1_in_a(T11))
P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A
U1_A(x1, x2)  =  U1_A(x2)

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:

P1_IN_A(T11) → P1_IN_A(T11)

The TRS R consists of the following rules:

p1_in_a(T4) → p1_out_a(T4)
p1_in_a(T11) → U1_a(T11, p1_in_a(T11))
U1_a(T11, p1_out_a(T11)) → p1_out_a(T11)

The argument filtering Pi contains the following mapping:
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a
U1_a(x1, x2)  =  U1_a(x2)
P1_IN_A(x1)  =  P1_IN_A

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:

P1_IN_A(T11) → P1_IN_A(T11)

R is empty.
The argument filtering Pi contains the following mapping:
P1_IN_A(x1)  =  P1_IN_A

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

(23) PiDPToQDPProof (SOUND 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:

P1_IN_AP1_IN_A

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 = P1_IN_A evaluates to t =P1_IN_A

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 P1_IN_A to P1_IN_A.



(26) FALSE