Left Termination of the query pattern log_in_2(g, a) w.r.t. the given Prolog program could successfully be proven:



↳ Prolog
  ↳ PrologToPiTRSProof

Clauses:

half(0, 0).
half(s(0), 0).
half(s(s(X)), s(Y)) :- half(X, Y).
log(0, s(0)).
log(s(X), s(Y)) :- ','(half(s(X), Z), log(Z, Y)).

Queries:

log(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:

log_in(s(X), s(Y)) → U2(X, Y, half_in(s(X), Z))
half_in(s(s(X)), s(Y)) → U1(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U1(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, Y, half_out(s(X), Z)) → U3(X, Y, log_in(Z, Y))
log_in(0, s(0)) → log_out(0, s(0))
U3(X, Y, log_out(Z, Y)) → log_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in(x1, x2)  =  log_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3)  =  U2(x3)
half_in(x1, x2)  =  half_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
half_out(x1, x2)  =  half_out(x2)
U3(x1, x2, x3)  =  U3(x3)
log_out(x1, x2)  =  log_out(x2)

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:

log_in(s(X), s(Y)) → U2(X, Y, half_in(s(X), Z))
half_in(s(s(X)), s(Y)) → U1(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U1(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, Y, half_out(s(X), Z)) → U3(X, Y, log_in(Z, Y))
log_in(0, s(0)) → log_out(0, s(0))
U3(X, Y, log_out(Z, Y)) → log_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in(x1, x2)  =  log_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3)  =  U2(x3)
half_in(x1, x2)  =  half_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
half_out(x1, x2)  =  half_out(x2)
U3(x1, x2, x3)  =  U3(x3)
log_out(x1, x2)  =  log_out(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:

LOG_IN(s(X), s(Y)) → U21(X, Y, half_in(s(X), Z))
LOG_IN(s(X), s(Y)) → HALF_IN(s(X), Z)
HALF_IN(s(s(X)), s(Y)) → U11(X, Y, half_in(X, Y))
HALF_IN(s(s(X)), s(Y)) → HALF_IN(X, Y)
U21(X, Y, half_out(s(X), Z)) → U31(X, Y, log_in(Z, Y))
U21(X, Y, half_out(s(X), Z)) → LOG_IN(Z, Y)

The TRS R consists of the following rules:

log_in(s(X), s(Y)) → U2(X, Y, half_in(s(X), Z))
half_in(s(s(X)), s(Y)) → U1(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U1(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, Y, half_out(s(X), Z)) → U3(X, Y, log_in(Z, Y))
log_in(0, s(0)) → log_out(0, s(0))
U3(X, Y, log_out(Z, Y)) → log_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in(x1, x2)  =  log_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3)  =  U2(x3)
half_in(x1, x2)  =  half_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
half_out(x1, x2)  =  half_out(x2)
U3(x1, x2, x3)  =  U3(x3)
log_out(x1, x2)  =  log_out(x2)
U31(x1, x2, x3)  =  U31(x3)
HALF_IN(x1, x2)  =  HALF_IN(x1)
U21(x1, x2, x3)  =  U21(x3)
U11(x1, x2, x3)  =  U11(x3)
LOG_IN(x1, x2)  =  LOG_IN(x1)

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:

LOG_IN(s(X), s(Y)) → U21(X, Y, half_in(s(X), Z))
LOG_IN(s(X), s(Y)) → HALF_IN(s(X), Z)
HALF_IN(s(s(X)), s(Y)) → U11(X, Y, half_in(X, Y))
HALF_IN(s(s(X)), s(Y)) → HALF_IN(X, Y)
U21(X, Y, half_out(s(X), Z)) → U31(X, Y, log_in(Z, Y))
U21(X, Y, half_out(s(X), Z)) → LOG_IN(Z, Y)

The TRS R consists of the following rules:

log_in(s(X), s(Y)) → U2(X, Y, half_in(s(X), Z))
half_in(s(s(X)), s(Y)) → U1(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U1(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, Y, half_out(s(X), Z)) → U3(X, Y, log_in(Z, Y))
log_in(0, s(0)) → log_out(0, s(0))
U3(X, Y, log_out(Z, Y)) → log_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in(x1, x2)  =  log_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3)  =  U2(x3)
half_in(x1, x2)  =  half_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
half_out(x1, x2)  =  half_out(x2)
U3(x1, x2, x3)  =  U3(x3)
log_out(x1, x2)  =  log_out(x2)
U31(x1, x2, x3)  =  U31(x3)
HALF_IN(x1, x2)  =  HALF_IN(x1)
U21(x1, x2, x3)  =  U21(x3)
U11(x1, x2, x3)  =  U11(x3)
LOG_IN(x1, x2)  =  LOG_IN(x1)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 3 less nodes.

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

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

HALF_IN(s(s(X)), s(Y)) → HALF_IN(X, Y)

The TRS R consists of the following rules:

log_in(s(X), s(Y)) → U2(X, Y, half_in(s(X), Z))
half_in(s(s(X)), s(Y)) → U1(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U1(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, Y, half_out(s(X), Z)) → U3(X, Y, log_in(Z, Y))
log_in(0, s(0)) → log_out(0, s(0))
U3(X, Y, log_out(Z, Y)) → log_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in(x1, x2)  =  log_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3)  =  U2(x3)
half_in(x1, x2)  =  half_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
half_out(x1, x2)  =  half_out(x2)
U3(x1, x2, x3)  =  U3(x3)
log_out(x1, x2)  =  log_out(x2)
HALF_IN(x1, x2)  =  HALF_IN(x1)

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
              ↳ PiDP

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

HALF_IN(s(s(X)), s(Y)) → HALF_IN(X, Y)

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

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
            ↳ AND
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ PiDP

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

HALF_IN(s(s(X))) → HALF_IN(X)

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: 



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

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

U21(X, Y, half_out(s(X), Z)) → LOG_IN(Z, Y)
LOG_IN(s(X), s(Y)) → U21(X, Y, half_in(s(X), Z))

The TRS R consists of the following rules:

log_in(s(X), s(Y)) → U2(X, Y, half_in(s(X), Z))
half_in(s(s(X)), s(Y)) → U1(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U1(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, Y, half_out(s(X), Z)) → U3(X, Y, log_in(Z, Y))
log_in(0, s(0)) → log_out(0, s(0))
U3(X, Y, log_out(Z, Y)) → log_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in(x1, x2)  =  log_in(x1)
s(x1)  =  s(x1)
U2(x1, x2, x3)  =  U2(x3)
half_in(x1, x2)  =  half_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
half_out(x1, x2)  =  half_out(x2)
U3(x1, x2, x3)  =  U3(x3)
log_out(x1, x2)  =  log_out(x2)
U21(x1, x2, x3)  =  U21(x3)
LOG_IN(x1, x2)  =  LOG_IN(x1)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof

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

U21(X, Y, half_out(s(X), Z)) → LOG_IN(Z, Y)
LOG_IN(s(X), s(Y)) → U21(X, Y, half_in(s(X), Z))

The TRS R consists of the following rules:

half_in(s(s(X)), s(Y)) → U1(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
U1(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in(0, 0) → half_out(0, 0)

The argument filtering Pi contains the following mapping:
s(x1)  =  s(x1)
half_in(x1, x2)  =  half_in(x1)
U1(x1, x2, x3)  =  U1(x3)
0  =  0
half_out(x1, x2)  =  half_out(x2)
U21(x1, x2, x3)  =  U21(x3)
LOG_IN(x1, x2)  =  LOG_IN(x1)

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
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof

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

LOG_IN(s(X)) → U21(half_in(s(X)))
U21(half_out(Z)) → LOG_IN(Z)

The TRS R consists of the following rules:

half_in(s(s(X))) → U1(half_in(X))
half_in(s(0)) → half_out(0)
U1(half_out(Y)) → half_out(s(Y))
half_in(0) → half_out(0)

The set Q consists of the following terms:

half_in(x0)
U1(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

half_in(s(0)) → half_out(0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1   
POL(LOG_IN(x1)) = 2·x1   
POL(U1(x1)) = 2·x1   
POL(U21(x1)) = 2·x1   
POL(half_in(x1)) = x1   
POL(half_out(x1)) = x1   
POL(s(x1)) = 2·x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof

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

LOG_IN(s(X)) → U21(half_in(s(X)))
U21(half_out(Z)) → LOG_IN(Z)

The TRS R consists of the following rules:

half_in(s(s(X))) → U1(half_in(X))
U1(half_out(Y)) → half_out(s(Y))
half_in(0) → half_out(0)

The set Q consists of the following terms:

half_in(x0)
U1(x0)

We have to consider all (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

half_in(s(s(X))) → U1(half_in(X))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0   
POL(LOG_IN(x1)) = 2·x1   
POL(U1(x1)) = 2 + x1   
POL(U21(x1)) = 2·x1   
POL(half_in(x1)) = x1   
POL(half_out(x1)) = x1   
POL(s(x1)) = 2 + x1   



↳ Prolog
  ↳ PrologToPiTRSProof
    ↳ PiTRS
      ↳ DependencyPairsProof
        ↳ PiDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ PiDP
              ↳ PiDP
                ↳ UsableRulesProof
                  ↳ PiDP
                    ↳ PiDPToQDPProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ RuleRemovalProof
                              ↳ QDP
                                ↳ DependencyGraphProof

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

LOG_IN(s(X)) → U21(half_in(s(X)))
U21(half_out(Z)) → LOG_IN(Z)

The TRS R consists of the following rules:

U1(half_out(Y)) → half_out(s(Y))
half_in(0) → half_out(0)

The set Q consists of the following terms:

half_in(x0)
U1(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.