Left Termination of the query pattern reverse_in_3(g, g, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

reverse([], X, X).
reverse(.(X, Y), Z, U) :- reverse(Y, Z, .(X, U)).

Queries:

reverse(g,g,a).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (b,b,f) (f,b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
  ↳ PrologToPiTRSProof

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

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x5)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN_GGA(.(X, Y), Z, U) → U1_GGA(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
REVERSE_IN_GGA(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))
REVERSE_IN_AGG(.(X, Y), Z, U) → U1_AGG(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
REVERSE_IN_AGG(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x5)
U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x5)
U1_AGG(x1, x2, x3, x4, x5)  =  U1_AGG(x5)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)
REVERSE_IN_GGA(x1, x2, x3)  =  REVERSE_IN_GGA(x1, x2)

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

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
  ↳ PrologToPiTRSProof

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

REVERSE_IN_GGA(.(X, Y), Z, U) → U1_GGA(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
REVERSE_IN_GGA(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))
REVERSE_IN_AGG(.(X, Y), Z, U) → U1_AGG(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
REVERSE_IN_AGG(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x5)
U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x5)
U1_AGG(x1, x2, x3, x4, x5)  =  U1_AGG(x5)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)
REVERSE_IN_GGA(x1, x2, x3)  =  REVERSE_IN_GGA(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
  ↳ PrologToPiTRSProof

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

REVERSE_IN_AGG(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x5)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
  ↳ PrologToPiTRSProof

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

REVERSE_IN_AGG(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Instantiation
  ↳ PrologToPiTRSProof

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

REVERSE_IN_AGG(Z, U) → REVERSE_IN_AGG(Z, .)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN_AGG(Z, U) → REVERSE_IN_AGG(Z, .) we obtained the following new rules:

REVERSE_IN_AGG(z0, .) → REVERSE_IN_AGG(z0, .)



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

REVERSE_IN_AGG(z0, .) → REVERSE_IN_AGG(z0, .)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

REVERSE_IN_AGG(z0, .) → REVERSE_IN_AGG(z0, .)

The TRS R consists of the following rules:none


s = REVERSE_IN_AGG(z0, .) evaluates to t =REVERSE_IN_AGG(z0, .)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from REVERSE_IN_AGG(z0, .) to REVERSE_IN_AGG(z0, .).




We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (b,b,f) (f,b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2)
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x3, x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x3, x4, x5)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2)
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x3, x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x3, x4, x5)


Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

REVERSE_IN_GGA(.(X, Y), Z, U) → U1_GGA(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
REVERSE_IN_GGA(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))
REVERSE_IN_AGG(.(X, Y), Z, U) → U1_AGG(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
REVERSE_IN_AGG(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2)
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x3, x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x3, x4, x5)
U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x3, x5)
U1_AGG(x1, x2, x3, x4, x5)  =  U1_AGG(x3, x4, x5)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)
REVERSE_IN_GGA(x1, x2, x3)  =  REVERSE_IN_GGA(x1, x2)

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

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof

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

REVERSE_IN_GGA(.(X, Y), Z, U) → U1_GGA(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
REVERSE_IN_GGA(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))
REVERSE_IN_AGG(.(X, Y), Z, U) → U1_AGG(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
REVERSE_IN_AGG(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2)
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x3, x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x3, x4, x5)
U1_GGA(x1, x2, x3, x4, x5)  =  U1_GGA(x3, x5)
U1_AGG(x1, x2, x3, x4, x5)  =  U1_AGG(x3, x4, x5)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)
REVERSE_IN_GGA(x1, x2, x3)  =  REVERSE_IN_GGA(x1, x2)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 3 less nodes.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof

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

REVERSE_IN_AGG(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))

The TRS R consists of the following rules:

reverse_in_gga([], X, X) → reverse_out_gga([], X, X)
reverse_in_gga(.(X, Y), Z, U) → U1_gga(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
reverse_in_agg([], X, X) → reverse_out_agg([], X, X)
reverse_in_agg(.(X, Y), Z, U) → U1_agg(X, Y, Z, U, reverse_in_agg(Y, Z, .(X, U)))
U1_agg(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_agg(.(X, Y), Z, U)
U1_gga(X, Y, Z, U, reverse_out_agg(Y, Z, .(X, U))) → reverse_out_gga(.(X, Y), Z, U)

The argument filtering Pi contains the following mapping:
reverse_in_gga(x1, x2, x3)  =  reverse_in_gga(x1, x2)
[]  =  []
reverse_out_gga(x1, x2, x3)  =  reverse_out_gga(x1, x2)
.(x1, x2)  =  .
U1_gga(x1, x2, x3, x4, x5)  =  U1_gga(x3, x5)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U1_agg(x1, x2, x3, x4, x5)  =  U1_agg(x3, x4, x5)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

We have to consider all (P,R,Pi)-chains
For (infinitary) constructor rewriting [30] we can delete all non-usable rules from R.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof

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

REVERSE_IN_AGG(.(X, Y), Z, U) → REVERSE_IN_AGG(Y, Z, .(X, U))

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

We have to consider all (P,R,Pi)-chains
Transforming (infinitary) constructor rewriting Pi-DP problem [30] into ordinary QDP problem [15] by application of Pi.

↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Instantiation

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

REVERSE_IN_AGG(Z, U) → REVERSE_IN_AGG(Z, .)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN_AGG(Z, U) → REVERSE_IN_AGG(Z, .) we obtained the following new rules:

REVERSE_IN_AGG(z0, .) → REVERSE_IN_AGG(z0, .)



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ PiDP
              ↳ UsableRulesProof
                ↳ PiDP
                  ↳ PiDPToQDPProof
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ NonTerminationProof

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

REVERSE_IN_AGG(z0, .) → REVERSE_IN_AGG(z0, .)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

REVERSE_IN_AGG(z0, .) → REVERSE_IN_AGG(z0, .)

The TRS R consists of the following rules:none


s = REVERSE_IN_AGG(z0, .) evaluates to t =REVERSE_IN_AGG(z0, .)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from REVERSE_IN_AGG(z0, .) to REVERSE_IN_AGG(z0, .).