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 NonTerminationProof (⇔)
    ↳12 FALSE
  ↳13 PrologToPiTRSProof (⇐)
    ↳14 PiTRS
    ↳15 DependencyPairsProof (⇔)
    ↳16 PiDP
    ↳17 DependencyGraphProof (⇔)
    ↳18 PiDP
    ↳19 UsableRulesProof (⇔)
    ↳20 PiDP
    ↳21 PiDPToQDPProof (⇐)
    ↳22 QDP
    ↳23 NonTerminationProof (⇔)
    ↳24 FALSE

(0) Obligation:

Clauses:

p(b).
p(a) :- p1(X).
p1(b).
p1(a) :- p1(X).

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

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)

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(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)

(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(a) → U1_G(p1_in_a(X))
P_IN_G(a) → P1_IN_A(X)
P1_IN_A(a) → U2_A(p1_in_a(X))
P1_IN_A(a) → P1_IN_A(X)

The TRS R consists of the following rules:

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1)  =  U1_G(x1)
P1_IN_A(x1)  =  P1_IN_A
U2_A(x1)  =  U2_A(x1)

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(a) → U1_G(p1_in_a(X))
P_IN_G(a) → P1_IN_A(X)
P1_IN_A(a) → U2_A(p1_in_a(X))
P1_IN_A(a) → P1_IN_A(X)

The TRS R consists of the following rules:

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1)  =  U1_G(x1)
P1_IN_A(x1)  =  P1_IN_A
U2_A(x1)  =  U2_A(x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

P1_IN_A(a) → P1_IN_A(X)

The TRS R consists of the following rules:

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
P1_IN_A(x1)  =  P1_IN_A

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:

P1_IN_A(a) → P1_IN_A(X)

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

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

(9) PiDPToQDPProof (SOUND 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:

P1_IN_A → P1_IN_A

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

(11) 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:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from P1_IN_A to P1_IN_A.



(12) FALSE

(13) 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)
p1_in: (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g(x1)
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(14) Obligation:

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

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g(x1)
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)

(15) 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(a) → U1_G(p1_in_a(X))
P_IN_G(a) → P1_IN_A(X)
P1_IN_A(a) → U2_A(p1_in_a(X))
P1_IN_A(a) → P1_IN_A(X)

The TRS R consists of the following rules:

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g(x1)
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1)  =  U1_G(x1)
P1_IN_A(x1)  =  P1_IN_A
U2_A(x1)  =  U2_A(x1)

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

(16) Obligation:

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

P_IN_G(a) → U1_G(p1_in_a(X))
P_IN_G(a) → P1_IN_A(X)
P1_IN_A(a) → U2_A(p1_in_a(X))
P1_IN_A(a) → P1_IN_A(X)

The TRS R consists of the following rules:

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g(x1)
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1)  =  U1_G(x1)
P1_IN_A(x1)  =  P1_IN_A
U2_A(x1)  =  U2_A(x1)

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(18) Obligation:

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

P1_IN_A(a) → P1_IN_A(X)

The TRS R consists of the following rules:

p_in_g(b) → p_out_g(b)
p_in_g(a) → U1_g(p1_in_a(X))
p1_in_a(b) → p1_out_a(b)
p1_in_a(a) → U2_a(p1_in_a(X))
U2_a(p1_out_a(X)) → p1_out_a(a)
U1_g(p1_out_a(X)) → p_out_g(a)

The argument filtering Pi contains the following mapping:
p_in_g(x1)  =  p_in_g(x1)
b  =  b
p_out_g(x1)  =  p_out_g(x1)
a  =  a
U1_g(x1)  =  U1_g(x1)
p1_in_a(x1)  =  p1_in_a
p1_out_a(x1)  =  p1_out_a(x1)
U2_a(x1)  =  U2_a(x1)
P1_IN_A(x1)  =  P1_IN_A

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

(19) UsableRulesProof (EQUIVALENT transformation)

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

(20) Obligation:

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

P1_IN_A(a) → P1_IN_A(X)

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

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

(21) PiDPToQDPProof (SOUND transformation)

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

(22) Obligation:

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

P1_IN_A → P1_IN_A

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

(23) 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.



(24) FALSE