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

0 Prolog
↳1 PrologToPrologProblemTransformerProof (⇐)
↳2 Prolog
    ↳3 PrologToPiTRSProof (⇐)
        ↳4 PiTRS
        ↳5 DependencyPairsProof (⇔)
        ↳6 PiDP
        ↳7 DependencyGraphProof (⇔)
        ↳8 AND
            ↳9 PiDP
                ↳10 UsableRulesProof (⇔)
                ↳11 PiDP
                ↳12 PiDPToQDPProof (⇔)
                ↳13 QDP
                ↳14 NonTerminationProof (⇔)
                ↳15 NO
            ↳16 PiDP
                ↳17 UsableRulesProof (⇔)
                ↳18 PiDP
                ↳19 PiDPToQDPProof (⇔)
                ↳20 QDP
                ↳21 NonTerminationProof (⇔)
                ↳22 NO
    ↳23 PrologToPiTRSProof (⇐)
        ↳24 PiTRS
        ↳25 DependencyPairsProof (⇔)
        ↳26 PiDP
        ↳27 DependencyGraphProof (⇔)
        ↳28 AND
            ↳29 PiDP
                ↳30 UsableRulesProof (⇔)
                ↳31 PiDP
                ↳32 PiDPToQDPProof (⇔)
                ↳33 QDP
                ↳34 NonTerminationProof (⇔)
                ↳35 NO
            ↳36 PiDP
                ↳37 UsableRulesProof (⇔)
                ↳38 PiDP
                ↳39 PiDPToQDPProof (⇔)
                ↳40 QDP
                ↳41 NonTerminationProof (⇔)
                ↳42 NO

(0) Obligation:

Clauses:

p(a).
p(X) :- p(a).
q(b).

Queries:

p(g).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

p6.
p6 :- p6.
p13.
p13 :- p13.
p1(a).
p1(a) :- p6.
p1(T3) :- p13.

Queries:

p1(g).

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

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

The argument filtering Pi contains the following mapping:
p1_in_g(x1)  =  p1_in_g(x1)
a  =  a
p1_out_g(x1)  =  p1_out_g
U3_g(x1)  =  U3_g(x1)
p6_in_  =  p6_in_
p6_out_  =  p6_out_
U1_(x1)  =  U1_(x1)
U4_g(x1, x2)  =  U4_g(x2)
p13_in_  =  p13_in_
p13_out_  =  p13_out_
U2_(x1)  =  U2_(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:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

The argument filtering Pi contains the following mapping:
p1_in_g(x1)  =  p1_in_g(x1)
a  =  a
p1_out_g(x1)  =  p1_out_g
U3_g(x1)  =  U3_g(x1)
p6_in_  =  p6_in_
p6_out_  =  p6_out_
U1_(x1)  =  U1_(x1)
U4_g(x1, x2)  =  U4_g(x2)
p13_in_  =  p13_in_
p13_out_  =  p13_out_
U2_(x1)  =  U2_(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:

P1_IN_G(a) → U3_G(p6_in_)
P1_IN_G(a) → P6_IN_
P6_IN_ → U1_1(p6_in_)
P6_IN_ → P6_IN_
P1_IN_G(T3) → U4_G(T3, p13_in_)
P1_IN_G(T3) → P13_IN_
P13_IN_ → U2_1(p13_in_)
P13_IN_ → P13_IN_

The TRS R consists of the following rules:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

The argument filtering Pi contains the following mapping:
p1_in_g(x1)  =  p1_in_g(x1)
a  =  a
p1_out_g(x1)  =  p1_out_g
U3_g(x1)  =  U3_g(x1)
p6_in_  =  p6_in_
p6_out_  =  p6_out_
U1_(x1)  =  U1_(x1)
U4_g(x1, x2)  =  U4_g(x2)
p13_in_  =  p13_in_
p13_out_  =  p13_out_
U2_(x1)  =  U2_(x1)
P1_IN_G(x1)  =  P1_IN_G(x1)
U3_G(x1)  =  U3_G(x1)
P6_IN_  =  P6_IN_
U1_1(x1)  =  U1_1(x1)
U4_G(x1, x2)  =  U4_G(x2)
P13_IN_  =  P13_IN_
U2_1(x1)  =  U2_1(x1)

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

(6) Obligation:

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

P1_IN_G(a) → U3_G(p6_in_)
P1_IN_G(a) → P6_IN_
P6_IN_ → U1_1(p6_in_)
P6_IN_ → P6_IN_
P1_IN_G(T3) → U4_G(T3, p13_in_)
P1_IN_G(T3) → P13_IN_
P13_IN_ → U2_1(p13_in_)
P13_IN_ → P13_IN_

The TRS R consists of the following rules:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

The argument filtering Pi contains the following mapping:
p1_in_g(x1)  =  p1_in_g(x1)
a  =  a
p1_out_g(x1)  =  p1_out_g
U3_g(x1)  =  U3_g(x1)
p6_in_  =  p6_in_
p6_out_  =  p6_out_
U1_(x1)  =  U1_(x1)
U4_g(x1, x2)  =  U4_g(x2)
p13_in_  =  p13_in_
p13_out_  =  p13_out_
U2_(x1)  =  U2_(x1)
P1_IN_G(x1)  =  P1_IN_G(x1)
U3_G(x1)  =  U3_G(x1)
P6_IN_  =  P6_IN_
U1_1(x1)  =  U1_1(x1)
U4_G(x1, x2)  =  U4_G(x2)
P13_IN_  =  P13_IN_
U2_1(x1)  =  U2_1(x1)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

P13_IN_ → P13_IN_

The TRS R consists of the following rules:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

The argument filtering Pi contains the following mapping:
p1_in_g(x1)  =  p1_in_g(x1)
a  =  a
p1_out_g(x1)  =  p1_out_g
U3_g(x1)  =  U3_g(x1)
p6_in_  =  p6_in_
p6_out_  =  p6_out_
U1_(x1)  =  U1_(x1)
U4_g(x1, x2)  =  U4_g(x2)
p13_in_  =  p13_in_
p13_out_  =  p13_out_
U2_(x1)  =  U2_(x1)
P13_IN_  =  P13_IN_

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

(10) UsableRulesProof (EQUIVALENT transformation)

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

(11) Obligation:

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

P13_IN_ → P13_IN_

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

(12) PiDPToQDPProof (EQUIVALENT transformation)

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

(13) Obligation:

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

P13_IN_ → P13_IN_

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

(14) 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 = P13_IN_ evaluates to t =P13_IN_

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 P13_IN_ to P13_IN_.



(15) NO

(16) Obligation:

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

P6_IN_ → P6_IN_

The TRS R consists of the following rules:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

The argument filtering Pi contains the following mapping:
p1_in_g(x1)  =  p1_in_g(x1)
a  =  a
p1_out_g(x1)  =  p1_out_g
U3_g(x1)  =  U3_g(x1)
p6_in_  =  p6_in_
p6_out_  =  p6_out_
U1_(x1)  =  U1_(x1)
U4_g(x1, x2)  =  U4_g(x2)
p13_in_  =  p13_in_
p13_out_  =  p13_out_
U2_(x1)  =  U2_(x1)
P6_IN_  =  P6_IN_

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

(17) UsableRulesProof (EQUIVALENT transformation)

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

(18) Obligation:

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

P6_IN_ → P6_IN_

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

(19) PiDPToQDPProof (EQUIVALENT transformation)

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

(20) Obligation:

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

P6_IN_ → P6_IN_

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

(21) 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 = P6_IN_ evaluates to t =P6_IN_

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 P6_IN_ to P6_IN_.



(22) NO

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

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(24) Obligation:

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

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

Pi is empty.

(25) 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_G(a) → U3_G(p6_in_)
P1_IN_G(a) → P6_IN_
P6_IN_ → U1_1(p6_in_)
P6_IN_ → P6_IN_
P1_IN_G(T3) → U4_G(T3, p13_in_)
P1_IN_G(T3) → P13_IN_
P13_IN_ → U2_1(p13_in_)
P13_IN_ → P13_IN_

The TRS R consists of the following rules:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

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

(26) Obligation:

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

P1_IN_G(a) → U3_G(p6_in_)
P1_IN_G(a) → P6_IN_
P6_IN_ → U1_1(p6_in_)
P6_IN_ → P6_IN_
P1_IN_G(T3) → U4_G(T3, p13_in_)
P1_IN_G(T3) → P13_IN_
P13_IN_ → U2_1(p13_in_)
P13_IN_ → P13_IN_

The TRS R consists of the following rules:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

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

(27) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(28) Complex Obligation (AND)

(29) Obligation:

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

P13_IN_ → P13_IN_

The TRS R consists of the following rules:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

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

(30) UsableRulesProof (EQUIVALENT transformation)

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

(31) Obligation:

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

P13_IN_ → P13_IN_

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

(32) PiDPToQDPProof (EQUIVALENT transformation)

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

(33) Obligation:

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

P13_IN_ → P13_IN_

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

(34) 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 = P13_IN_ evaluates to t =P13_IN_

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 P13_IN_ to P13_IN_.



(35) NO

(36) Obligation:

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

P6_IN_ → P6_IN_

The TRS R consists of the following rules:

p1_in_g(a) → p1_out_g(a)
p1_in_g(a) → U3_g(p6_in_)
p6_in_ → p6_out_
p6_in_ → U1_(p6_in_)
U1_(p6_out_) → p6_out_
U3_g(p6_out_) → p1_out_g(a)
p1_in_g(T3) → U4_g(T3, p13_in_)
p13_in_ → p13_out_
p13_in_ → U2_(p13_in_)
U2_(p13_out_) → p13_out_
U4_g(T3, p13_out_) → p1_out_g(T3)

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

(37) UsableRulesProof (EQUIVALENT transformation)

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

(38) Obligation:

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

P6_IN_ → P6_IN_

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

(39) PiDPToQDPProof (EQUIVALENT transformation)

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

(40) Obligation:

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

P6_IN_ → P6_IN_

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

(41) 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 = P6_IN_ evaluates to t =P6_IN_

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 P6_IN_ to P6_IN_.



(42) NO