Left Termination of the query pattern goal_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(a).
p(X) :- p(Y).
q(b).
goal(X) :- ','(p(X), q(X)).

Queries:

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

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
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_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
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_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g

(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:

GOAL_IN_G(X) → U2_G(X, p_in_g(X))
GOAL_IN_G(X) → P_IN_G(X)
P_IN_G(X) → U1_G(X, p_in_a(Y))
P_IN_G(X) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U2_G(X, p_out_g(X)) → U3_G(X, q_in_g(X))
U2_G(X, p_out_g(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
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_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x1, x2)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
P_IN_A(x1)  =  P_IN_A
U1_A(x1, x2)  =  U1_A(x2)
U3_G(x1, x2)  =  U3_G(x2)
Q_IN_G(x1)  =  Q_IN_G(x1)

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

(4) Obligation:

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

GOAL_IN_G(X) → U2_G(X, p_in_g(X))
GOAL_IN_G(X) → P_IN_G(X)
P_IN_G(X) → U1_G(X, p_in_a(Y))
P_IN_G(X) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U2_G(X, p_out_g(X)) → U3_G(X, q_in_g(X))
U2_G(X, p_out_g(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
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_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x1, x2)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x2)
P_IN_A(x1)  =  P_IN_A
U1_A(x1, x2)  =  U1_A(x2)
U3_G(x1, x2)  =  U3_G(x2)
Q_IN_G(x1)  =  Q_IN_G(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 7 less nodes.

(6) Obligation:

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

P_IN_A(X) → P_IN_A(Y)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
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_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g
goal_out_g(x1)  =  goal_out_g
P_IN_A(x1)  =  P_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:

P_IN_A(X) → P_IN_A(Y)

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

P_IN_A → P_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 = P_IN_A evaluates to t =P_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 P_IN_A to P_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:
goal_in: (b)
p_in: (b) (f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(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:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(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:

GOAL_IN_G(X) → U2_G(X, p_in_g(X))
GOAL_IN_G(X) → P_IN_G(X)
P_IN_G(X) → U1_G(X, p_in_a(Y))
P_IN_G(X) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U2_G(X, p_out_g(X)) → U3_G(X, q_in_g(X))
U2_G(X, p_out_g(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(x1)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x1, x2)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
P_IN_A(x1)  =  P_IN_A
U1_A(x1, x2)  =  U1_A(x2)
U3_G(x1, x2)  =  U3_G(x1, x2)
Q_IN_G(x1)  =  Q_IN_G(x1)

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

(16) Obligation:

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

GOAL_IN_G(X) → U2_G(X, p_in_g(X))
GOAL_IN_G(X) → P_IN_G(X)
P_IN_G(X) → U1_G(X, p_in_a(Y))
P_IN_G(X) → P_IN_A(Y)
P_IN_A(X) → U1_A(X, p_in_a(Y))
P_IN_A(X) → P_IN_A(Y)
U2_G(X, p_out_g(X)) → U3_G(X, q_in_g(X))
U2_G(X, p_out_g(X)) → Q_IN_G(X)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(x1)
GOAL_IN_G(x1)  =  GOAL_IN_G(x1)
U2_G(x1, x2)  =  U2_G(x1, x2)
P_IN_G(x1)  =  P_IN_G(x1)
U1_G(x1, x2)  =  U1_G(x1, x2)
P_IN_A(x1)  =  P_IN_A
U1_A(x1, x2)  =  U1_A(x2)
U3_G(x1, x2)  =  U3_G(x1, x2)
Q_IN_G(x1)  =  Q_IN_G(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 7 less nodes.

(18) Obligation:

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

P_IN_A(X) → P_IN_A(Y)

The TRS R consists of the following rules:

goal_in_g(X) → U2_g(X, p_in_g(X))
p_in_g(a) → p_out_g(a)
p_in_g(X) → U1_g(X, p_in_a(Y))
p_in_a(a) → p_out_a(a)
p_in_a(X) → U1_a(X, p_in_a(Y))
U1_a(X, p_out_a(Y)) → p_out_a(X)
U1_g(X, p_out_a(Y)) → p_out_g(X)
U2_g(X, p_out_g(X)) → U3_g(X, q_in_g(X))
q_in_g(b) → q_out_g(b)
U3_g(X, q_out_g(X)) → goal_out_g(X)

The argument filtering Pi contains the following mapping:
goal_in_g(x1)  =  goal_in_g(x1)
U2_g(x1, x2)  =  U2_g(x1, x2)
p_in_g(x1)  =  p_in_g(x1)
a  =  a
p_out_g(x1)  =  p_out_g(x1)
U1_g(x1, x2)  =  U1_g(x1, x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a
U1_a(x1, x2)  =  U1_a(x2)
U3_g(x1, x2)  =  U3_g(x1, x2)
q_in_g(x1)  =  q_in_g(x1)
b  =  b
q_out_g(x1)  =  q_out_g(x1)
goal_out_g(x1)  =  goal_out_g(x1)
P_IN_A(x1)  =  P_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:

P_IN_A(X) → P_IN_A(Y)

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

P_IN_A → P_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 = P_IN_A evaluates to t =P_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 P_IN_A to P_IN_A.



(24) FALSE