Left Termination of the query pattern q_in_1(a) w.r.t. the given Prolog program could not be shown:

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

(0) Obligation:

Clauses:

q(X) :- ','(p(X), !).
p(0).
p(s(X)) :- p(X).

Queries:

q(a).

(1) CutEliminatorProof (SOUND transformation)

Eliminated all cuts by simply ignoring them[PROLOG].

(2) Obligation:

Clauses:

q(X) :- p(X).
p(0).
p(s(X)) :- p(X).

Queries:

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

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)

(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:

Q_IN_A(X) → U1_A(X, p_in_a(X))
Q_IN_A(X) → P_IN_A(X)
P_IN_A(s(X)) → U2_A(X, p_in_a(X))
P_IN_A(s(X)) → P_IN_A(X)

The TRS R consists of the following rules:

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)
Q_IN_A(x1)  =  Q_IN_A
U1_A(x1, x2)  =  U1_A(x2)
P_IN_A(x1)  =  P_IN_A
U2_A(x1, x2)  =  U2_A(x2)

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

(6) Obligation:

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

Q_IN_A(X) → U1_A(X, p_in_a(X))
Q_IN_A(X) → P_IN_A(X)
P_IN_A(s(X)) → U2_A(X, p_in_a(X))
P_IN_A(s(X)) → P_IN_A(X)

The TRS R consists of the following rules:

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)
Q_IN_A(x1)  =  Q_IN_A
U1_A(x1, x2)  =  U1_A(x2)
P_IN_A(x1)  =  P_IN_A
U2_A(x1, x2)  =  U2_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 3 less nodes.

(8) Obligation:

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

P_IN_A(s(X)) → P_IN_A(X)

The TRS R consists of the following rules:

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)
P_IN_A(x1)  =  P_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:

P_IN_A(s(X)) → P_IN_A(X)

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

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

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)

(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:

Q_IN_A(X) → U1_A(X, p_in_a(X))
Q_IN_A(X) → P_IN_A(X)
P_IN_A(s(X)) → U2_A(X, p_in_a(X))
P_IN_A(s(X)) → P_IN_A(X)

The TRS R consists of the following rules:

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)
Q_IN_A(x1)  =  Q_IN_A
U1_A(x1, x2)  =  U1_A(x2)
P_IN_A(x1)  =  P_IN_A
U2_A(x1, x2)  =  U2_A(x2)

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

(18) Obligation:

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

Q_IN_A(X) → U1_A(X, p_in_a(X))
Q_IN_A(X) → P_IN_A(X)
P_IN_A(s(X)) → U2_A(X, p_in_a(X))
P_IN_A(s(X)) → P_IN_A(X)

The TRS R consists of the following rules:

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)
Q_IN_A(x1)  =  Q_IN_A
U1_A(x1, x2)  =  U1_A(x2)
P_IN_A(x1)  =  P_IN_A
U2_A(x1, x2)  =  U2_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 3 less nodes.

(20) Obligation:

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

P_IN_A(s(X)) → P_IN_A(X)

The TRS R consists of the following rules:

q_in_a(X) → U1_a(X, p_in_a(X))
p_in_a(0) → p_out_a(0)
p_in_a(s(X)) → U2_a(X, p_in_a(X))
U2_a(X, p_out_a(X)) → p_out_a(s(X))
U1_a(X, p_out_a(X)) → q_out_a(X)

The argument filtering Pi contains the following mapping:
q_in_a(x1)  =  q_in_a
U1_a(x1, x2)  =  U1_a(x2)
p_in_a(x1)  =  p_in_a
p_out_a(x1)  =  p_out_a(x1)
U2_a(x1, x2)  =  U2_a(x2)
q_out_a(x1)  =  q_out_a(x1)
s(x1)  =  s(x1)
P_IN_A(x1)  =  P_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:

P_IN_A(s(X)) → P_IN_A(X)

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

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



(26) FALSE