Left Termination of the query pattern less_in_2(g, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

less(0, s(X)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

less(g,a).

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

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)

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:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)


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

LESS_IN_GA(s(X), s(Y)) → U1_GA(X, Y, less_in_aa(X, Y))
LESS_IN_GA(s(X), s(Y)) → LESS_IN_AA(X, Y)
LESS_IN_AA(s(X), s(Y)) → U1_AA(X, Y, less_in_aa(X, Y))
LESS_IN_AA(s(X), s(Y)) → LESS_IN_AA(X, Y)

The TRS R consists of the following rules:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)
LESS_IN_AA(x1, x2)  =  LESS_IN_AA
U1_AA(x1, x2, x3)  =  U1_AA(x3)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)

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:

LESS_IN_GA(s(X), s(Y)) → U1_GA(X, Y, less_in_aa(X, Y))
LESS_IN_GA(s(X), s(Y)) → LESS_IN_AA(X, Y)
LESS_IN_AA(s(X), s(Y)) → U1_AA(X, Y, less_in_aa(X, Y))
LESS_IN_AA(s(X), s(Y)) → LESS_IN_AA(X, Y)

The TRS R consists of the following rules:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)
LESS_IN_AA(x1, x2)  =  LESS_IN_AA
U1_AA(x1, x2, x3)  =  U1_AA(x3)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)

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:

LESS_IN_AA(s(X), s(Y)) → LESS_IN_AA(X, Y)

The TRS R consists of the following rules:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)
LESS_IN_AA(x1, x2)  =  LESS_IN_AA

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:

LESS_IN_AA(s(X), s(Y)) → LESS_IN_AA(X, Y)

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

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
                      ↳ NonTerminationProof
  ↳ PrologToPiTRSProof

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

LESS_IN_AA → LESS_IN_AA

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:

LESS_IN_AA → LESS_IN_AA

The TRS R consists of the following rules:none


s = LESS_IN_AA evaluates to t =LESS_IN_AA

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 LESS_IN_AA to LESS_IN_AA.




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

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1, x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)

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:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1, x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)


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

LESS_IN_GA(s(X), s(Y)) → U1_GA(X, Y, less_in_aa(X, Y))
LESS_IN_GA(s(X), s(Y)) → LESS_IN_AA(X, Y)
LESS_IN_AA(s(X), s(Y)) → U1_AA(X, Y, less_in_aa(X, Y))
LESS_IN_AA(s(X), s(Y)) → LESS_IN_AA(X, Y)

The TRS R consists of the following rules:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1, x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)
LESS_IN_AA(x1, x2)  =  LESS_IN_AA
U1_AA(x1, x2, x3)  =  U1_AA(x3)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)

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:

LESS_IN_GA(s(X), s(Y)) → U1_GA(X, Y, less_in_aa(X, Y))
LESS_IN_GA(s(X), s(Y)) → LESS_IN_AA(X, Y)
LESS_IN_AA(s(X), s(Y)) → U1_AA(X, Y, less_in_aa(X, Y))
LESS_IN_AA(s(X), s(Y)) → LESS_IN_AA(X, Y)

The TRS R consists of the following rules:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1, x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)
LESS_IN_AA(x1, x2)  =  LESS_IN_AA
U1_AA(x1, x2, x3)  =  U1_AA(x3)
U1_GA(x1, x2, x3)  =  U1_GA(x3)
LESS_IN_GA(x1, x2)  =  LESS_IN_GA(x1)

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:

LESS_IN_AA(s(X), s(Y)) → LESS_IN_AA(X, Y)

The TRS R consists of the following rules:

less_in_ga(0, s(X)) → less_out_ga(0, s(X))
less_in_ga(s(X), s(Y)) → U1_ga(X, Y, less_in_aa(X, Y))
less_in_aa(0, s(X)) → less_out_aa(0, s(X))
less_in_aa(s(X), s(Y)) → U1_aa(X, Y, less_in_aa(X, Y))
U1_aa(X, Y, less_out_aa(X, Y)) → less_out_aa(s(X), s(Y))
U1_ga(X, Y, less_out_aa(X, Y)) → less_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
less_in_ga(x1, x2)  =  less_in_ga(x1)
0  =  0
less_out_ga(x1, x2)  =  less_out_ga(x1, x2)
s(x1)  =  s
U1_ga(x1, x2, x3)  =  U1_ga(x3)
less_in_aa(x1, x2)  =  less_in_aa
less_out_aa(x1, x2)  =  less_out_aa(x1, x2)
U1_aa(x1, x2, x3)  =  U1_aa(x3)
LESS_IN_AA(x1, x2)  =  LESS_IN_AA

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:

LESS_IN_AA(s(X), s(Y)) → LESS_IN_AA(X, Y)

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

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
                      ↳ NonTerminationProof

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

LESS_IN_AA → LESS_IN_AA

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:

LESS_IN_AA → LESS_IN_AA

The TRS R consists of the following rules:none


s = LESS_IN_AA evaluates to t =LESS_IN_AA

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 LESS_IN_AA to LESS_IN_AA.