Left Termination of the query pattern p_in_1(g) w.r.t. the given Prolog program could not be shown:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇔)
    ↳10 QDP
      ↳11 Instantiation (⇔)
        ↳12 QDP
      ↳13 Instantiation (⇔)
        ↳14 QDP
        ↳15 NonTerminationProof (⇔)
        ↳16 FALSE
  ↳17 PrologToPiTRSProof (⇐)
    ↳18 PiTRS
    ↳19 DependencyPairsProof (⇔)
    ↳20 PiDP
    ↳21 DependencyGraphProof (⇔)
    ↳22 PiDP
    ↳23 UsableRulesProof (⇔)
    ↳24 PiDP
    ↳25 PiDPToQDPProof (⇔)
    ↳26 QDP
      ↳27 Instantiation (⇔)
        ↳28 QDP
      ↳29 Instantiation (⇔)
        ↳30 QDP
        ↳31 NonTerminationProof (⇔)
        ↳32 FALSE

(0) Obligation:

Clauses:

p(a).
p(X) :- p(a).
q(b).

Queries:

p(g).

(1) PrologToPiTRSProof (SOUND transformation)

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

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)

(3) 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:

P_IN_G(X) → U1_G(X, p_in_g(a))
P_IN_G(X) → P_IN_G(a)

The TRS R consists of the following rules:

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(4) Obligation:

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

P_IN_G(X) → U1_G(X, p_in_g(a))
P_IN_G(X) → P_IN_G(a)

The TRS R consists of the following rules:

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

P_IN_G(X) → P_IN_G(a)

The TRS R consists of the following rules:

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g
U1_g(x1, x2)  =  U1_g(x2)
P_IN_G(x1)  =  P_IN_G(x1)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

P_IN_G(X) → P_IN_G(a)

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

(9) PiDPToQDPProof (EQUIVALENT transformation)

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

(10) Obligation:

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

P_IN_G(X) → P_IN_G(a)

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

(11) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule P_IN_G(X) → P_IN_G(a) we obtained the following new rules [LPAR04]:

P_IN_G(a) → P_IN_G(a)

(12) Obligation:

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

P_IN_G(a) → P_IN_G(a)

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

(13) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule P_IN_G(X) → P_IN_G(a) we obtained the following new rules [LPAR04]:

P_IN_G(a) → P_IN_G(a)

(14) Obligation:

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

P_IN_G(a) → P_IN_G(a)

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

(15) 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 = P_IN_G(a) evaluates to t =P_IN_G(a)

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 P_IN_G(a) to P_IN_G(a).



(16) FALSE

(17) PrologToPiTRSProof (SOUND transformation)

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

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(18) Obligation:

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

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

Pi is empty.

(19) 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:

P_IN_G(X) → U1_G(X, p_in_g(a))
P_IN_G(X) → P_IN_G(a)

The TRS R consists of the following rules:

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

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

(20) Obligation:

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

P_IN_G(X) → U1_G(X, p_in_g(a))
P_IN_G(X) → P_IN_G(a)

The TRS R consists of the following rules:

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

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

(21) DependencyGraphProof (EQUIVALENT transformation)

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

(22) Obligation:

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

P_IN_G(X) → P_IN_G(a)

The TRS R consists of the following rules:

p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_g(a))
U1_g(X, p_out_g(a)) → p_out_g(X)

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

(23) UsableRulesProof (EQUIVALENT transformation)

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

(24) Obligation:

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

P_IN_G(X) → P_IN_G(a)

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

(25) PiDPToQDPProof (EQUIVALENT transformation)

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

(26) Obligation:

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

P_IN_G(X) → P_IN_G(a)

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

(27) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule P_IN_G(X) → P_IN_G(a) we obtained the following new rules [LPAR04]:

P_IN_G(a) → P_IN_G(a)

(28) Obligation:

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

P_IN_G(a) → P_IN_G(a)

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

(29) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule P_IN_G(X) → P_IN_G(a) we obtained the following new rules [LPAR04]:

P_IN_G(a) → P_IN_G(a)

(30) Obligation:

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

P_IN_G(a) → P_IN_G(a)

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

(31) 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 = P_IN_G(a) evaluates to t =P_IN_G(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 P_IN_G(a) to P_IN_G(a).



(32) FALSE