Left Termination of the query pattern reverse_in_2(a, g) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

reverse(X1s, X2s) :- reverse(X1s, [], X2s).
reverse([], Xs, Xs).
reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).

Queries:

reverse(a,g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (f,b)
reverse_in: (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_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)

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_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)


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_AG(X1s, X2s) → U1_AG(X1s, X2s, reverse_in_agg(X1s, [], X2s))
REVERSE_IN_AG(X1s, X2s) → REVERSE_IN_AGG(X1s, [], X2s)
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → U2_AGG(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x5)
U1_AG(x1, x2, x3)  =  U1_AG(x3)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

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_AG(X1s, X2s) → U1_AG(X1s, X2s, reverse_in_agg(X1s, [], X2s))
REVERSE_IN_AG(X1s, X2s) → REVERSE_IN_AGG(X1s, [], X2s)
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → U2_AGG(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x5)
U1_AG(x1, x2, x3)  =  U1_AG(x3)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

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, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
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, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(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(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN_AGG(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys) we obtained the following new rules:

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)



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

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

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1) we obtained the following new rules:

REVERSE_IN_AGG(.(.(z0)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)



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

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

REVERSE_IN_AGG(.(.(z0)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)

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)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)

The TRS R consists of the following rules:none


s = REVERSE_IN_AGG(.(.(z0)), z1) evaluates to t =REVERSE_IN_AGG(.(.(.(z0))), z1)

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)), z1) to REVERSE_IN_AGG(.(.(.(z0))), z1).




We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
reverse_in: (f,b)
reverse_in: (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_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)

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_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)


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_AG(X1s, X2s) → U1_AG(X1s, X2s, reverse_in_agg(X1s, [], X2s))
REVERSE_IN_AG(X1s, X2s) → REVERSE_IN_AGG(X1s, [], X2s)
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → U2_AGG(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x3, x4, x5)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

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_AG(X1s, X2s) → U1_AG(X1s, X2s, reverse_in_agg(X1s, [], X2s))
REVERSE_IN_AG(X1s, X2s) → REVERSE_IN_AGG(X1s, [], X2s)
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → U2_AGG(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x3, x4, x5)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

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, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
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, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(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(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN_AGG(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys) we obtained the following new rules:

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)



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

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

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1) we obtained the following new rules:

REVERSE_IN_AGG(.(.(z0)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)



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

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

REVERSE_IN_AGG(.(.(z0)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)

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)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)

The TRS R consists of the following rules:none


s = REVERSE_IN_AGG(.(.(z0)), z1) evaluates to t =REVERSE_IN_AGG(.(.(.(z0))), z1)

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)), z1) to REVERSE_IN_AGG(.(.(.(z0))), z1).