Left Termination of the query pattern max_in_3(a, a, g) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ 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,a,g).

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

max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)

The argument filtering Pi contains the following mapping:
max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
less_in_ag(x1, x2)  =  less_in_ag(x2)
s(x1)  =  s(x1)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U3_ag(x1, x2, x3)  =  U3_ag(x3)
max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof

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

max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)

The argument filtering Pi contains the following mapping:
max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
less_in_ag(x1, x2)  =  less_in_ag(x2)
s(x1)  =  s(x1)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U3_ag(x1, x2, x3)  =  U3_ag(x3)
max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
U2_aag(x1, x2, x3)  =  U2_aag(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_AAG(X, Y, X) → U1_AAG(X, Y, less_in_ag(Y, X))
MAX_IN_AAG(X, Y, X) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U3_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
MAX_IN_AAG(X, Y, Y) → U2_AAG(X, Y, less_in_ag(X, s(Y)))
MAX_IN_AAG(X, Y, Y) → LESS_IN_AG(X, s(Y))

The TRS R consists of the following rules:

max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)

The argument filtering Pi contains the following mapping:
max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
less_in_ag(x1, x2)  =  less_in_ag(x2)
s(x1)  =  s(x1)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U3_ag(x1, x2, x3)  =  U3_ag(x3)
max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)
U3_AG(x1, x2, x3)  =  U3_AG(x3)
U2_AAG(x1, x2, x3)  =  U2_AAG(x2, x3)
U1_AAG(x1, x2, x3)  =  U1_AAG(x1, x3)
MAX_IN_AAG(x1, x2, x3)  =  MAX_IN_AAG(x3)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)

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

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

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

MAX_IN_AAG(X, Y, X) → U1_AAG(X, Y, less_in_ag(Y, X))
MAX_IN_AAG(X, Y, X) → LESS_IN_AG(Y, X)
LESS_IN_AG(s(X), s(Y)) → U3_AG(X, Y, less_in_ag(X, Y))
LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)
MAX_IN_AAG(X, Y, Y) → U2_AAG(X, Y, less_in_ag(X, s(Y)))
MAX_IN_AAG(X, Y, Y) → LESS_IN_AG(X, s(Y))

The TRS R consists of the following rules:

max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)

The argument filtering Pi contains the following mapping:
max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
less_in_ag(x1, x2)  =  less_in_ag(x2)
s(x1)  =  s(x1)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U3_ag(x1, x2, x3)  =  U3_ag(x3)
max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)
U3_AG(x1, x2, x3)  =  U3_AG(x3)
U2_AAG(x1, x2, x3)  =  U2_AAG(x2, x3)
U1_AAG(x1, x2, x3)  =  U1_AAG(x1, x3)
MAX_IN_AAG(x1, x2, x3)  =  MAX_IN_AAG(x3)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(x2)

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

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

LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)

The TRS R consists of the following rules:

max_in_aag(X, Y, X) → U1_aag(X, Y, less_in_ag(Y, X))
less_in_ag(0, s(X)) → less_out_ag(0, s(X))
less_in_ag(s(X), s(Y)) → U3_ag(X, Y, less_in_ag(X, Y))
U3_ag(X, Y, less_out_ag(X, Y)) → less_out_ag(s(X), s(Y))
U1_aag(X, Y, less_out_ag(Y, X)) → max_out_aag(X, Y, X)
max_in_aag(X, Y, Y) → U2_aag(X, Y, less_in_ag(X, s(Y)))
U2_aag(X, Y, less_out_ag(X, s(Y))) → max_out_aag(X, Y, Y)

The argument filtering Pi contains the following mapping:
max_in_aag(x1, x2, x3)  =  max_in_aag(x3)
U1_aag(x1, x2, x3)  =  U1_aag(x1, x3)
less_in_ag(x1, x2)  =  less_in_ag(x2)
s(x1)  =  s(x1)
less_out_ag(x1, x2)  =  less_out_ag(x1)
U3_ag(x1, x2, x3)  =  U3_ag(x3)
max_out_aag(x1, x2, x3)  =  max_out_aag(x1, x2)
U2_aag(x1, x2, x3)  =  U2_aag(x2, x3)
LESS_IN_AG(x1, x2)  =  LESS_IN_AG(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

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

LESS_IN_AG(s(X), s(Y)) → LESS_IN_AG(X, Y)

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

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

LESS_IN_AG(s(Y)) → LESS_IN_AG(Y)

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: