Left Termination of the query pattern max_in_3(a, g, a) w.r.t. the given Prolog program could not be shown:



↳ Prolog
  ↳ PrologToPiTRSProof
  ↳ PrologToPiTRSProof

Clauses:

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

Queries:

max(a,g,a).

We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out
U1(x1, x2, x3)  =  U1(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:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out
U1(x1, x2, x3)  =  U1(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:

MAX_IN(X, Y, Y) → U21(X, Y, less_in(X, s(Y)))
MAX_IN(X, Y, Y) → LESS_IN(X, s(Y))
LESS_IN(s(X), s(Y)) → U31(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
MAX_IN(X, Y, X) → U11(X, Y, less_in(Y, X))
MAX_IN(X, Y, X) → LESS_IN(Y, X)

The TRS R consists of the following rules:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out
U1(x1, x2, x3)  =  U1(x3)
MAX_IN(x1, x2, x3)  =  MAX_IN(x2)
LESS_IN(x1, x2)  =  LESS_IN
U31(x1, x2, x3)  =  U31(x3)
U21(x1, x2, x3)  =  U21(x3)
U11(x1, x2, x3)  =  U11(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:

MAX_IN(X, Y, Y) → U21(X, Y, less_in(X, s(Y)))
MAX_IN(X, Y, Y) → LESS_IN(X, s(Y))
LESS_IN(s(X), s(Y)) → U31(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
MAX_IN(X, Y, X) → U11(X, Y, less_in(Y, X))
MAX_IN(X, Y, X) → LESS_IN(Y, X)

The TRS R consists of the following rules:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out
U1(x1, x2, x3)  =  U1(x3)
MAX_IN(x1, x2, x3)  =  MAX_IN(x2)
LESS_IN(x1, x2)  =  LESS_IN
U31(x1, x2, x3)  =  U31(x3)
U21(x1, x2, x3)  =  U21(x3)
U11(x1, x2, x3)  =  U11(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.

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

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

The TRS R consists of the following rules:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out
U1(x1, x2, x3)  =  U1(x3)
LESS_IN(x1, x2)  =  LESS_IN

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(s(X), s(Y)) → LESS_IN(X, Y)

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

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 → LESS_IN

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 → LESS_IN

The TRS R consists of the following rules:none


s = LESS_IN evaluates to t =LESS_IN

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 to LESS_IN.




We use the technique of [30].Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x2, x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out(x2)
U1(x1, x2, x3)  =  U1(x2, 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:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x2, x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out(x2)
U1(x1, x2, x3)  =  U1(x2, 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:

MAX_IN(X, Y, Y) → U21(X, Y, less_in(X, s(Y)))
MAX_IN(X, Y, Y) → LESS_IN(X, s(Y))
LESS_IN(s(X), s(Y)) → U31(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
MAX_IN(X, Y, X) → U11(X, Y, less_in(Y, X))
MAX_IN(X, Y, X) → LESS_IN(Y, X)

The TRS R consists of the following rules:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x2, x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out(x2)
U1(x1, x2, x3)  =  U1(x2, x3)
MAX_IN(x1, x2, x3)  =  MAX_IN(x2)
LESS_IN(x1, x2)  =  LESS_IN
U31(x1, x2, x3)  =  U31(x3)
U21(x1, x2, x3)  =  U21(x2, x3)
U11(x1, x2, x3)  =  U11(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:

MAX_IN(X, Y, Y) → U21(X, Y, less_in(X, s(Y)))
MAX_IN(X, Y, Y) → LESS_IN(X, s(Y))
LESS_IN(s(X), s(Y)) → U31(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
MAX_IN(X, Y, X) → U11(X, Y, less_in(Y, X))
MAX_IN(X, Y, X) → LESS_IN(Y, X)

The TRS R consists of the following rules:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x2, x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out(x2)
U1(x1, x2, x3)  =  U1(x2, x3)
MAX_IN(x1, x2, x3)  =  MAX_IN(x2)
LESS_IN(x1, x2)  =  LESS_IN
U31(x1, x2, x3)  =  U31(x3)
U21(x1, x2, x3)  =  U21(x2, x3)
U11(x1, x2, x3)  =  U11(x2, x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 1 SCC with 5 less nodes.

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

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

The TRS R consists of the following rules:

max_in(X, Y, Y) → U2(X, Y, less_in(X, s(Y)))
less_in(s(X), s(Y)) → U3(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U3(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U2(X, Y, less_out(X, s(Y))) → max_out(X, Y, Y)
max_in(X, Y, X) → U1(X, Y, less_in(Y, X))
U1(X, Y, less_out(Y, X)) → max_out(X, Y, X)

The argument filtering Pi contains the following mapping:
max_in(x1, x2, x3)  =  max_in(x2)
U2(x1, x2, x3)  =  U2(x2, x3)
less_in(x1, x2)  =  less_in
s(x1)  =  s(x1)
U3(x1, x2, x3)  =  U3(x3)
0  =  0
less_out(x1, x2)  =  less_out(x1)
max_out(x1, x2, x3)  =  max_out(x2)
U1(x1, x2, x3)  =  U1(x2, x3)
LESS_IN(x1, x2)  =  LESS_IN

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(s(X), s(Y)) → LESS_IN(X, Y)

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

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 → LESS_IN

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 → LESS_IN

The TRS R consists of the following rules:none


s = LESS_IN evaluates to t =LESS_IN

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 to LESS_IN.