Left Termination of the query pattern minus_in_3(g, g, 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:

minus(X, Y, Z) :- ','(f(X, 0), ','(!, =(Z, 0))).
f(X, Y) :- ','(!, =(X, Y)).
f(X, Y) :- f(X, Y).
=(X, X).

Queries:

minus(g,g,a).

(1) CutEliminatorProof (SOUND transformation)

Eliminated all cuts by simply ignoring them[PROLOG].

(2) Obligation:

Clauses:

minus(X, Y, Z) :- ','(f(X, 0), =(Z, 0)).
f(X, Y) :- =(X, Y).
f(X, Y) :- f(X, Y).
=(X, X).

Queries:

minus(g,g,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:
minus_in: (b,b,f)
f_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg
f_out_gg(x1, x2)  =  f_out_gg
U4_gg(x1, x2, x3)  =  U4_gg(x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg
f_out_gg(x1, x2)  =  f_out_gg
U4_gg(x1, x2, x3)  =  U4_gg(x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x3)

(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:

MINUS_IN_GGA(X, Y, Z) → U1_GGA(X, Y, Z, f_in_gg(X, 0))
MINUS_IN_GGA(X, Y, Z) → F_IN_GG(X, 0)
F_IN_GG(X, Y) → U3_GG(X, Y, =_in_gg(X, Y))
F_IN_GG(X, Y) → =_IN_GG(X, Y)
F_IN_GG(X, Y) → U4_GG(X, Y, f_in_gg(X, Y))
F_IN_GG(X, Y) → F_IN_GG(X, Y)
U1_GGA(X, Y, Z, f_out_gg(X, 0)) → U2_GGA(X, Y, Z, =_in_ag(Z, 0))
U1_GGA(X, Y, Z, f_out_gg(X, 0)) → =_IN_AG(Z, 0)

The TRS R consists of the following rules:

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg
f_out_gg(x1, x2)  =  f_out_gg
U4_gg(x1, x2, x3)  =  U4_gg(x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x3)
MINUS_IN_GGA(x1, x2, x3)  =  MINUS_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x4)
F_IN_GG(x1, x2)  =  F_IN_GG(x1, x2)
U3_GG(x1, x2, x3)  =  U3_GG(x3)
=_IN_GG(x1, x2)  =  =_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x3)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x4)
=_IN_AG(x1, x2)  =  =_IN_AG(x2)

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

(6) Obligation:

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

MINUS_IN_GGA(X, Y, Z) → U1_GGA(X, Y, Z, f_in_gg(X, 0))
MINUS_IN_GGA(X, Y, Z) → F_IN_GG(X, 0)
F_IN_GG(X, Y) → U3_GG(X, Y, =_in_gg(X, Y))
F_IN_GG(X, Y) → =_IN_GG(X, Y)
F_IN_GG(X, Y) → U4_GG(X, Y, f_in_gg(X, Y))
F_IN_GG(X, Y) → F_IN_GG(X, Y)
U1_GGA(X, Y, Z, f_out_gg(X, 0)) → U2_GGA(X, Y, Z, =_in_ag(Z, 0))
U1_GGA(X, Y, Z, f_out_gg(X, 0)) → =_IN_AG(Z, 0)

The TRS R consists of the following rules:

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg
f_out_gg(x1, x2)  =  f_out_gg
U4_gg(x1, x2, x3)  =  U4_gg(x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x3)
MINUS_IN_GGA(x1, x2, x3)  =  MINUS_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x4)
F_IN_GG(x1, x2)  =  F_IN_GG(x1, x2)
U3_GG(x1, x2, x3)  =  U3_GG(x3)
=_IN_GG(x1, x2)  =  =_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x3)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x4)
=_IN_AG(x1, x2)  =  =_IN_AG(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 7 less nodes.

(8) Obligation:

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

F_IN_GG(X, Y) → F_IN_GG(X, Y)

The TRS R consists of the following rules:

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg
f_out_gg(x1, x2)  =  f_out_gg
U4_gg(x1, x2, x3)  =  U4_gg(x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x3)
F_IN_GG(x1, x2)  =  F_IN_GG(x1, x2)

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:

F_IN_GG(X, Y) → F_IN_GG(X, Y)

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

(11) PiDPToQDPProof (EQUIVALENT 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:

F_IN_GG(X, Y) → F_IN_GG(X, Y)

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 = F_IN_GG(X, Y) evaluates to t =F_IN_GG(X, Y)

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 F_IN_GG(X, Y) to F_IN_GG(X, Y).



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

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x2, x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg(x1, x2)
f_out_gg(x1, x2)  =  f_out_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1, x2)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x1, x2, x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x2, x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg(x1, x2)
f_out_gg(x1, x2)  =  f_out_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1, x2)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x1, x2, x3)

(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:

MINUS_IN_GGA(X, Y, Z) → U1_GGA(X, Y, Z, f_in_gg(X, 0))
MINUS_IN_GGA(X, Y, Z) → F_IN_GG(X, 0)
F_IN_GG(X, Y) → U3_GG(X, Y, =_in_gg(X, Y))
F_IN_GG(X, Y) → =_IN_GG(X, Y)
F_IN_GG(X, Y) → U4_GG(X, Y, f_in_gg(X, Y))
F_IN_GG(X, Y) → F_IN_GG(X, Y)
U1_GGA(X, Y, Z, f_out_gg(X, 0)) → U2_GGA(X, Y, Z, =_in_ag(Z, 0))
U1_GGA(X, Y, Z, f_out_gg(X, 0)) → =_IN_AG(Z, 0)

The TRS R consists of the following rules:

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x2, x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg(x1, x2)
f_out_gg(x1, x2)  =  f_out_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1, x2)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x1, x2, x3)
MINUS_IN_GGA(x1, x2, x3)  =  MINUS_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x1, x2, x4)
F_IN_GG(x1, x2)  =  F_IN_GG(x1, x2)
U3_GG(x1, x2, x3)  =  U3_GG(x1, x2, x3)
=_IN_GG(x1, x2)  =  =_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x1, x2, x3)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x1, x2, x4)
=_IN_AG(x1, x2)  =  =_IN_AG(x2)

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

(18) Obligation:

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

MINUS_IN_GGA(X, Y, Z) → U1_GGA(X, Y, Z, f_in_gg(X, 0))
MINUS_IN_GGA(X, Y, Z) → F_IN_GG(X, 0)
F_IN_GG(X, Y) → U3_GG(X, Y, =_in_gg(X, Y))
F_IN_GG(X, Y) → =_IN_GG(X, Y)
F_IN_GG(X, Y) → U4_GG(X, Y, f_in_gg(X, Y))
F_IN_GG(X, Y) → F_IN_GG(X, Y)
U1_GGA(X, Y, Z, f_out_gg(X, 0)) → U2_GGA(X, Y, Z, =_in_ag(Z, 0))
U1_GGA(X, Y, Z, f_out_gg(X, 0)) → =_IN_AG(Z, 0)

The TRS R consists of the following rules:

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x2, x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg(x1, x2)
f_out_gg(x1, x2)  =  f_out_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1, x2)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x1, x2, x3)
MINUS_IN_GGA(x1, x2, x3)  =  MINUS_IN_GGA(x1, x2)
U1_GGA(x1, x2, x3, x4)  =  U1_GGA(x1, x2, x4)
F_IN_GG(x1, x2)  =  F_IN_GG(x1, x2)
U3_GG(x1, x2, x3)  =  U3_GG(x1, x2, x3)
=_IN_GG(x1, x2)  =  =_IN_GG(x1, x2)
U4_GG(x1, x2, x3)  =  U4_GG(x1, x2, x3)
U2_GGA(x1, x2, x3, x4)  =  U2_GGA(x1, x2, x4)
=_IN_AG(x1, x2)  =  =_IN_AG(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 7 less nodes.

(20) Obligation:

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

F_IN_GG(X, Y) → F_IN_GG(X, Y)

The TRS R consists of the following rules:

minus_in_gga(X, Y, Z) → U1_gga(X, Y, Z, f_in_gg(X, 0))
f_in_gg(X, Y) → U3_gg(X, Y, =_in_gg(X, Y))
=_in_gg(X, X) → =_out_gg(X, X)
U3_gg(X, Y, =_out_gg(X, Y)) → f_out_gg(X, Y)
f_in_gg(X, Y) → U4_gg(X, Y, f_in_gg(X, Y))
U4_gg(X, Y, f_out_gg(X, Y)) → f_out_gg(X, Y)
U1_gga(X, Y, Z, f_out_gg(X, 0)) → U2_gga(X, Y, Z, =_in_ag(Z, 0))
=_in_ag(X, X) → =_out_ag(X, X)
U2_gga(X, Y, Z, =_out_ag(Z, 0)) → minus_out_gga(X, Y, Z)

The argument filtering Pi contains the following mapping:
minus_in_gga(x1, x2, x3)  =  minus_in_gga(x1, x2)
U1_gga(x1, x2, x3, x4)  =  U1_gga(x1, x2, x4)
f_in_gg(x1, x2)  =  f_in_gg(x1, x2)
U3_gg(x1, x2, x3)  =  U3_gg(x1, x2, x3)
=_in_gg(x1, x2)  =  =_in_gg(x1, x2)
=_out_gg(x1, x2)  =  =_out_gg(x1, x2)
f_out_gg(x1, x2)  =  f_out_gg(x1, x2)
U4_gg(x1, x2, x3)  =  U4_gg(x1, x2, x3)
0  =  0
U2_gga(x1, x2, x3, x4)  =  U2_gga(x1, x2, x4)
=_in_ag(x1, x2)  =  =_in_ag(x2)
=_out_ag(x1, x2)  =  =_out_ag(x1, x2)
minus_out_gga(x1, x2, x3)  =  minus_out_gga(x1, x2, x3)
F_IN_GG(x1, x2)  =  F_IN_GG(x1, x2)

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:

F_IN_GG(X, Y) → F_IN_GG(X, Y)

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

(23) PiDPToQDPProof (EQUIVALENT 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:

F_IN_GG(X, Y) → F_IN_GG(X, Y)

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 = F_IN_GG(X, Y) evaluates to t =F_IN_GG(X, Y)

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 F_IN_GG(X, Y) to F_IN_GG(X, Y).



(26) FALSE