Left Termination of the query pattern plus_in_3(g, a, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

plus(0, Y, Y).
plus(s(X), Y, Z) :- plus(X, s(Y), Z).

Queries:

plus(g,a,a).

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

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)

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:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)


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

PLUS_IN_GAA(s(X), Y, Z) → U1_GAA(X, Y, Z, plus_in_aga(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)
PLUS_IN_AGA(s(X), Y, Z) → U1_AGA(X, Y, Z, plus_in_aga(X, s(Y), Z))
PLUS_IN_AGA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
PLUS_IN_AGA(x1, x2, x3)  =  PLUS_IN_AGA(x2)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_AGA(x1, x2, x3, x4)  =  U1_AGA(x4)

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:

PLUS_IN_GAA(s(X), Y, Z) → U1_GAA(X, Y, Z, plus_in_aga(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)
PLUS_IN_AGA(s(X), Y, Z) → U1_AGA(X, Y, Z, plus_in_aga(X, s(Y), Z))
PLUS_IN_AGA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
PLUS_IN_AGA(x1, x2, x3)  =  PLUS_IN_AGA(x2)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_AGA(x1, x2, x3, x4)  =  U1_AGA(x4)

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:

PLUS_IN_AGA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x4)
PLUS_IN_AGA(x1, x2, x3)  =  PLUS_IN_AGA(x2)

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:

PLUS_IN_AGA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)

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

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:

PLUS_IN_AGA(Y) → PLUS_IN_AGA(s)

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

PLUS_IN_AGA(s) → PLUS_IN_AGA(s)



↳ 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:

PLUS_IN_AGA(s) → PLUS_IN_AGA(s)

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:

PLUS_IN_AGA(s) → PLUS_IN_AGA(s)

The TRS R consists of the following rules:none


s = PLUS_IN_AGA(s) evaluates to t =PLUS_IN_AGA(s)

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 PLUS_IN_AGA(s) to PLUS_IN_AGA(s).




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

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x2, x4)

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:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x2, x4)


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

PLUS_IN_GAA(s(X), Y, Z) → U1_GAA(X, Y, Z, plus_in_aga(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)
PLUS_IN_AGA(s(X), Y, Z) → U1_AGA(X, Y, Z, plus_in_aga(X, s(Y), Z))
PLUS_IN_AGA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x2, x4)
PLUS_IN_AGA(x1, x2, x3)  =  PLUS_IN_AGA(x2)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_AGA(x1, x2, x3, x4)  =  U1_AGA(x2, x4)

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:

PLUS_IN_GAA(s(X), Y, Z) → U1_GAA(X, Y, Z, plus_in_aga(X, s(Y), Z))
PLUS_IN_GAA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)
PLUS_IN_AGA(s(X), Y, Z) → U1_AGA(X, Y, Z, plus_in_aga(X, s(Y), Z))
PLUS_IN_AGA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x2, x4)
PLUS_IN_AGA(x1, x2, x3)  =  PLUS_IN_AGA(x2)
U1_GAA(x1, x2, x3, x4)  =  U1_GAA(x4)
PLUS_IN_GAA(x1, x2, x3)  =  PLUS_IN_GAA(x1)
U1_AGA(x1, x2, x3, x4)  =  U1_AGA(x2, x4)

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:

PLUS_IN_AGA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)

The TRS R consists of the following rules:

plus_in_gaa(0, Y, Y) → plus_out_gaa(0, Y, Y)
plus_in_gaa(s(X), Y, Z) → U1_gaa(X, Y, Z, plus_in_aga(X, s(Y), Z))
plus_in_aga(0, Y, Y) → plus_out_aga(0, Y, Y)
plus_in_aga(s(X), Y, Z) → U1_aga(X, Y, Z, plus_in_aga(X, s(Y), Z))
U1_aga(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_aga(s(X), Y, Z)
U1_gaa(X, Y, Z, plus_out_aga(X, s(Y), Z)) → plus_out_gaa(s(X), Y, Z)

The argument filtering Pi contains the following mapping:
plus_in_gaa(x1, x2, x3)  =  plus_in_gaa(x1)
0  =  0
plus_out_gaa(x1, x2, x3)  =  plus_out_gaa(x1)
s(x1)  =  s
U1_gaa(x1, x2, x3, x4)  =  U1_gaa(x4)
plus_in_aga(x1, x2, x3)  =  plus_in_aga(x2)
plus_out_aga(x1, x2, x3)  =  plus_out_aga(x1, x2, x3)
U1_aga(x1, x2, x3, x4)  =  U1_aga(x2, x4)
PLUS_IN_AGA(x1, x2, x3)  =  PLUS_IN_AGA(x2)

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:

PLUS_IN_AGA(s(X), Y, Z) → PLUS_IN_AGA(X, s(Y), Z)

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

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:

PLUS_IN_AGA(Y) → PLUS_IN_AGA(s)

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

PLUS_IN_AGA(s) → PLUS_IN_AGA(s)



↳ 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:

PLUS_IN_AGA(s) → PLUS_IN_AGA(s)

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:

PLUS_IN_AGA(s) → PLUS_IN_AGA(s)

The TRS R consists of the following rules:none


s = PLUS_IN_AGA(s) evaluates to t =PLUS_IN_AGA(s)

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 PLUS_IN_AGA(s) to PLUS_IN_AGA(s).