Left Termination proof of /tmp/tmpNZjLz0/log.pl

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

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

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_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2) = log_in_ga(x1)
0 = 0
log_out_ga(x1, x2) = log_out_ga(x2)
s(x1) = s(x1)
U2_ga(x1, x2, x3) = U2_ga(x3)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
U3_ga(x1, x2, x3) = U3_ga(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:

LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))
LOG_IN_GA(s(X), s(Y)) → HALF_IN_GA(s(X), Z)
HALF_IN_GA(s(s(X)), s(Y)) → U1_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GA(X, Y, half_out_ga(s(X), Z)) → U3_GA(X, Y, log_in_ga(Z, Y))
U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2) = log_in_ga(x1)
0 = 0
log_out_ga(x1, x2) = log_out_ga(x2)
s(x1) = s(x1)
U2_ga(x1, x2, x3) = U2_ga(x3)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
U3_ga(x1, x2, x3) = U3_ga(x3)
LOG_IN_GA(x1, x2) = LOG_IN_GA(x1)
HALF_IN_GA(x1, x2) = HALF_IN_GA(x1)
U3_GA(x1, x2, x3) = U3_GA(x3)
U1_GA(x1, x2, x3) = U1_GA(x3)
U2_GA(x1, x2, x3) = U2_GA(x3)

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_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))
LOG_IN_GA(s(X), s(Y)) → HALF_IN_GA(s(X), Z)
HALF_IN_GA(s(s(X)), s(Y)) → U1_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GA(X, Y, half_out_ga(s(X), Z)) → U3_GA(X, Y, log_in_ga(Z, Y))
U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2) = log_in_ga(x1)
0 = 0
log_out_ga(x1, x2) = log_out_ga(x2)
s(x1) = s(x1)
U2_ga(x1, x2, x3) = U2_ga(x3)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
U3_ga(x1, x2, x3) = U3_ga(x3)
LOG_IN_GA(x1, x2) = LOG_IN_GA(x1)
HALF_IN_GA(x1, x2) = HALF_IN_GA(x1)
U3_GA(x1, x2, x3) = U3_GA(x3)
U1_GA(x1, x2, x3) = U1_GA(x3)
U2_GA(x1, x2, x3) = U2_GA(x3)

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_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2) = log_in_ga(x1)
0 = 0
log_out_ga(x1, x2) = log_out_ga(x2)
s(x1) = s(x1)
U2_ga(x1, x2, x3) = U2_ga(x3)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
U3_ga(x1, x2, x3) = U3_ga(x3)
HALF_IN_GA(x1, x2) = HALF_IN_GA(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_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)

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

HALF_IN_GA(s(s(X))) → HALF_IN_GA(X)
The graph contains the following edges 1 > 1

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))
U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)

The TRS R consists of the following rules:

log_in_ga(0, s(0)) → log_out_ga(0, s(0))
log_in_ga(s(X), s(Y)) → U2_ga(X, Y, half_in_ga(s(X), Z))
half_in_ga(0, 0) → half_out_ga(0, 0)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ga(X, Y, half_out_ga(s(X), Z)) → U3_ga(X, Y, log_in_ga(Z, Y))
U3_ga(X, Y, log_out_ga(Z, Y)) → log_out_ga(s(X), s(Y))

The argument filtering Pi contains the following mapping:
log_in_ga(x1, x2) = log_in_ga(x1)
0 = 0
log_out_ga(x1, x2) = log_out_ga(x2)
s(x1) = s(x1)
U2_ga(x1, x2, x3) = U2_ga(x3)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
U3_ga(x1, x2, x3) = U3_ga(x3)
LOG_IN_GA(x1, x2) = LOG_IN_GA(x1)
U2_GA(x1, x2, x3) = U2_GA(x3)

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:

LOG_IN_GA(s(X), s(Y)) → U2_GA(X, Y, half_in_ga(s(X), Z))
U2_GA(X, Y, half_out_ga(s(X), Z)) → LOG_IN_GA(Z, Y)

The TRS R consists of the following rules:

half_in_ga(s(0), 0) → half_out_ga(s(0), 0)
half_in_ga(s(s(X)), s(Y)) → U1_ga(X, Y, half_in_ga(X, Y))
U1_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
half_in_ga(0, 0) → half_out_ga(0, 0)

The argument filtering Pi contains the following mapping:
0 = 0
s(x1) = s(x1)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U1_ga(x1, x2, x3) = U1_ga(x3)
LOG_IN_GA(x1, x2) = LOG_IN_GA(x1)
U2_GA(x1, x2, x3) = U2_GA(x3)

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:

U2_GA(half_out_ga(Z)) → LOG_IN_GA(Z)
LOG_IN_GA(s(X)) → U2_GA(half_in_ga(s(X)))

The TRS R consists of the following rules:

half_in_ga(s(0)) → half_out_ga(0)
half_in_ga(s(s(X))) → U1_ga(half_in_ga(X))
U1_ga(half_out_ga(Y)) → half_out_ga(s(Y))
half_in_ga(0) → half_out_ga(0)

The set Q consists of the following terms:

half_in_ga(x0)
U1_ga(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_ga(s(0)) → half_out_ga(0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 2
POL(LOG_IN_GA(x₁)) = x₁
POL(U1_ga(x₁)) = 2·x₁
POL(U2_GA(x₁)) = x₁
POL(half_in_ga(x₁)) = x₁
POL(half_out_ga(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:

U2_GA(half_out_ga(Z)) → LOG_IN_GA(Z)
LOG_IN_GA(s(X)) → U2_GA(half_in_ga(s(X)))

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U1_ga(half_in_ga(X))
U1_ga(half_out_ga(Y)) → half_out_ga(s(Y))
half_in_ga(0) → half_out_ga(0)

The set Q consists of the following terms:

half_in_ga(x0)
U1_ga(x0)

LOG_IN_GA(s(X)) → U2_GA(half_in_ga(s(X)))

Strictly oriented rules of the TRS R:

half_in_ga(s(s(X))) → U1_ga(half_in_ga(X))

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(LOG_IN_GA(x₁)) = 2·x₁
POL(U1_ga(x₁)) = 2 + 2·x₁
POL(U2_GA(x₁)) = x₁
POL(half_in_ga(x₁)) = x₁
POL(half_out_ga(x₁)) = 2·x₁
POL(s(x₁)) = 1 + 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:

U2_GA(half_out_ga(Z)) → LOG_IN_GA(Z)

The TRS R consists of the following rules:

U1_ga(half_out_ga(Y)) → half_out_ga(s(Y))
half_in_ga(0) → half_out_ga(0)

The set Q consists of the following terms:

half_in_ga(x0)
U1_ga(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.