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

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

(0) Obligation:

Clauses:

a :- b.
a :- e.
b :- c.
c :- d.
d :- b.
e :- f.
f :- g.
g :- e.

Queries:

a().

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

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

Pi is empty.

(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:

A_IN_ → U1_1(b_in_)
A_IN_ → B_IN_
B_IN_ → U3_1(c_in_)
B_IN_ → C_IN_
C_IN_ → U4_1(d_in_)
C_IN_ → D_IN_
D_IN_ → U5_1(b_in_)
D_IN_ → B_IN_
A_IN_ → U2_1(e_in_)
A_IN_ → E_IN_
E_IN_ → U6_1(f_in_)
E_IN_ → F_IN_
F_IN_ → U7_1(g_in_)
F_IN_ → G_IN_
G_IN_ → U8_1(e_in_)
G_IN_ → E_IN_

The TRS R consists of the following rules:

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

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

(4) Obligation:

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

A_IN_ → U1_1(b_in_)
A_IN_ → B_IN_
B_IN_ → U3_1(c_in_)
B_IN_ → C_IN_
C_IN_ → U4_1(d_in_)
C_IN_ → D_IN_
D_IN_ → U5_1(b_in_)
D_IN_ → B_IN_
A_IN_ → U2_1(e_in_)
A_IN_ → E_IN_
E_IN_ → U6_1(f_in_)
E_IN_ → F_IN_
F_IN_ → U7_1(g_in_)
F_IN_ → G_IN_
G_IN_ → U8_1(e_in_)
G_IN_ → E_IN_

The TRS R consists of the following rules:

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 10 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

E_IN_ → F_IN_
F_IN_ → G_IN_
G_IN_ → E_IN_

The TRS R consists of the following rules:

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

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

(8) UsableRulesProof (EQUIVALENT transformation)

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

(9) Obligation:

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

E_IN_ → F_IN_
F_IN_ → G_IN_
G_IN_ → E_IN_

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

(10) PiDPToQDPProof (EQUIVALENT transformation)

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

(11) Obligation:

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

E_IN_ → F_IN_
F_IN_ → G_IN_
G_IN_ → E_IN_

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

(12) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F_IN_ evaluates to t =F_IN_

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

F_IN_ → G_IN_
with rule F_IN_ → G_IN_ at position [] and matcher [ ]

G_IN_ → E_IN_
with rule G_IN_ → E_IN_ at position [] and matcher [ ]

E_IN_ → F_IN_
with rule E_IN_ → F_IN_

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(13) FALSE

(14) Obligation:

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

B_IN_ → C_IN_
C_IN_ → D_IN_
D_IN_ → B_IN_

The TRS R consists of the following rules:

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

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

(15) UsableRulesProof (EQUIVALENT transformation)

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

(16) Obligation:

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

B_IN_ → C_IN_
C_IN_ → D_IN_
D_IN_ → B_IN_

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

(17) PiDPToQDPProof (EQUIVALENT transformation)

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

(18) Obligation:

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

B_IN_ → C_IN_
C_IN_ → D_IN_
D_IN_ → B_IN_

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

(19) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = C_IN_ evaluates to t =C_IN_

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

C_IN_ → D_IN_
with rule C_IN_ → D_IN_ at position [] and matcher [ ]

D_IN_ → B_IN_
with rule D_IN_ → B_IN_ at position [] and matcher [ ]

B_IN_ → C_IN_
with rule B_IN_ → C_IN_

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(20) FALSE

(21) PrologToPiTRSProof (SOUND transformation)

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

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(22) Obligation:

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

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

Pi is empty.

(23) 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:

A_IN_ → U1_1(b_in_)
A_IN_ → B_IN_
B_IN_ → U3_1(c_in_)
B_IN_ → C_IN_
C_IN_ → U4_1(d_in_)
C_IN_ → D_IN_
D_IN_ → U5_1(b_in_)
D_IN_ → B_IN_
A_IN_ → U2_1(e_in_)
A_IN_ → E_IN_
E_IN_ → U6_1(f_in_)
E_IN_ → F_IN_
F_IN_ → U7_1(g_in_)
F_IN_ → G_IN_
G_IN_ → U8_1(e_in_)
G_IN_ → E_IN_

The TRS R consists of the following rules:

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

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

(24) Obligation:

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

A_IN_ → U1_1(b_in_)
A_IN_ → B_IN_
B_IN_ → U3_1(c_in_)
B_IN_ → C_IN_
C_IN_ → U4_1(d_in_)
C_IN_ → D_IN_
D_IN_ → U5_1(b_in_)
D_IN_ → B_IN_
A_IN_ → U2_1(e_in_)
A_IN_ → E_IN_
E_IN_ → U6_1(f_in_)
E_IN_ → F_IN_
F_IN_ → U7_1(g_in_)
F_IN_ → G_IN_
G_IN_ → U8_1(e_in_)
G_IN_ → E_IN_

The TRS R consists of the following rules:

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 10 less nodes.

(26) Complex Obligation (AND)

(27) Obligation:

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

E_IN_ → F_IN_
F_IN_ → G_IN_
G_IN_ → E_IN_

The TRS R consists of the following rules:

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

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

(28) UsableRulesProof (EQUIVALENT transformation)

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

(29) Obligation:

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

E_IN_ → F_IN_
F_IN_ → G_IN_
G_IN_ → E_IN_

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

(30) PiDPToQDPProof (EQUIVALENT transformation)

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

(31) Obligation:

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

E_IN_ → F_IN_
F_IN_ → G_IN_
G_IN_ → E_IN_

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

(32) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F_IN_ evaluates to t =F_IN_

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

F_IN_ → G_IN_
with rule F_IN_ → G_IN_ at position [] and matcher [ ]

G_IN_ → E_IN_
with rule G_IN_ → E_IN_ at position [] and matcher [ ]

E_IN_ → F_IN_
with rule E_IN_ → F_IN_

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(33) FALSE

(34) Obligation:

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

B_IN_ → C_IN_
C_IN_ → D_IN_
D_IN_ → B_IN_

The TRS R consists of the following rules:

a_in_ → U1_(b_in_)
b_in_ → U3_(c_in_)
c_in_ → U4_(d_in_)
d_in_ → U5_(b_in_)
U5_(b_out_) → d_out_
U4_(d_out_) → c_out_
U3_(c_out_) → b_out_
U1_(b_out_) → a_out_
a_in_ → U2_(e_in_)
e_in_ → U6_(f_in_)
f_in_ → U7_(g_in_)
g_in_ → U8_(e_in_)
U8_(e_out_) → g_out_
U7_(g_out_) → f_out_
U6_(f_out_) → e_out_
U2_(e_out_) → a_out_

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

(35) UsableRulesProof (EQUIVALENT transformation)

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

(36) Obligation:

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

B_IN_ → C_IN_
C_IN_ → D_IN_
D_IN_ → B_IN_

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

(37) PiDPToQDPProof (EQUIVALENT transformation)

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

(38) Obligation:

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

B_IN_ → C_IN_
C_IN_ → D_IN_
D_IN_ → B_IN_

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

(39) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = C_IN_ evaluates to t =C_IN_

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

C_IN_ → D_IN_
with rule C_IN_ → D_IN_ at position [] and matcher [ ]

D_IN_ → B_IN_
with rule D_IN_ → B_IN_ at position [] and matcher [ ]

B_IN_ → C_IN_
with rule B_IN_ → C_IN_

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(40) FALSE