Left Termination proof of /tmp/tmpLeyke2/log2a-oi.pl

Left Termination of the query pattern log2_in_2(a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

log2(X, Y) :- log2(X, 0, Y).
log2(0, I, I).
log2(s(0), I, I).
log2(s(s(X)), I, Y) :- ','(half(s(s(X)), X1), log2(X1, s(I), Y)).
half(0, 0).
half(s(0), 0).
half(s(s(X)), s(Y)) :- half(X, Y).

Queries:

log2(a,g).

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
log2_in: (f,b)
log2_in: (f,b,b) (b,b,b)
half_in: (f,f) (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

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:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

LOG2_IN_AG(X, Y) → U1_AG(X, Y, log2_in_agg(X, 0, Y))
LOG2_IN_AG(X, Y) → LOG2_IN_AGG(X, 0, Y)
LOG2_IN_AGG(s(s(X)), I, Y) → U2_AGG(X, I, Y, half_in_aa(s(s(X)), X1))
LOG2_IN_AGG(s(s(X)), I, Y) → HALF_IN_AA(s(s(X)), X1)
HALF_IN_AA(s(s(X)), s(Y)) → U4_AA(X, Y, half_in_aa(X, Y))
HALF_IN_AA(s(s(X)), s(Y)) → HALF_IN_AA(X, Y)
U2_AGG(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_AGG(X, I, Y, log2_in_ggg(X1, s(I), Y))
U2_AGG(X, I, Y, half_out_aa(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)
LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)), X1))
LOG2_IN_GGG(s(s(X)), I, Y) → HALF_IN_GA(s(s(X)), X1)
HALF_IN_GA(s(s(X)), s(Y)) → U4_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_GGG(X, I, Y, log2_in_ggg(X1, s(I), Y))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ag(x1, x2) = log2_in_ag(x2)
U1_ag(x1, x2, x3) = U1_ag(x3)
log2_in_agg(x1, x2, x3) = log2_in_agg(x2, x3)
log2_out_agg(x1, x2, x3) = log2_out_agg(x1)
U2_agg(x1, x2, x3, x4) = U2_agg(x2, x3, x4)
half_in_aa(x1, x2) = half_in_aa
half_out_aa(x1, x2) = half_out_aa(x1, x2)
U4_aa(x1, x2, x3) = U4_aa(x3)
s(x1) = s(x1)
U3_agg(x1, x2, x3, x4) = U3_agg(x1, x4)
log2_in_ggg(x1, x2, x3) = log2_in_ggg(x1, x2, x3)
0 = 0
log2_out_ggg(x1, x2, x3) = log2_out_ggg
U2_ggg(x1, x2, x3, x4) = U2_ggg(x2, x3, x4)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U4_ga(x1, x2, x3) = U4_ga(x3)
U3_ggg(x1, x2, x3, x4) = U3_ggg(x4)
log2_out_ag(x1, x2) = log2_out_ag(x1)
U3_AGG(x1, x2, x3, x4) = U3_AGG(x1, x4)
LOG2_IN_AGG(x1, x2, x3) = LOG2_IN_AGG(x2, x3)
LOG2_IN_GGG(x1, x2, x3) = LOG2_IN_GGG(x1, x2, x3)
U3_GGG(x1, x2, x3, x4) = U3_GGG(x4)
HALF_IN_GA(x1, x2) = HALF_IN_GA(x1)
U2_AGG(x1, x2, x3, x4) = U2_AGG(x2, x3, x4)
U4_GA(x1, x2, x3) = U4_GA(x3)
U2_GGG(x1, x2, x3, x4) = U2_GGG(x2, x3, x4)
HALF_IN_AA(x1, x2) = HALF_IN_AA
U1_AG(x1, x2, x3) = U1_AG(x3)
U4_AA(x1, x2, x3) = U4_AA(x3)
LOG2_IN_AG(x1, x2) = LOG2_IN_AG(x2)

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:

LOG2_IN_AG(X, Y) → U1_AG(X, Y, log2_in_agg(X, 0, Y))
LOG2_IN_AG(X, Y) → LOG2_IN_AGG(X, 0, Y)
LOG2_IN_AGG(s(s(X)), I, Y) → U2_AGG(X, I, Y, half_in_aa(s(s(X)), X1))
LOG2_IN_AGG(s(s(X)), I, Y) → HALF_IN_AA(s(s(X)), X1)
HALF_IN_AA(s(s(X)), s(Y)) → U4_AA(X, Y, half_in_aa(X, Y))
HALF_IN_AA(s(s(X)), s(Y)) → HALF_IN_AA(X, Y)
U2_AGG(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_AGG(X, I, Y, log2_in_ggg(X1, s(I), Y))
U2_AGG(X, I, Y, half_out_aa(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)
LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)), X1))
LOG2_IN_GGG(s(s(X)), I, Y) → HALF_IN_GA(s(s(X)), X1)
HALF_IN_GA(s(s(X)), s(Y)) → U4_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_GGG(X, I, Y, log2_in_ggg(X1, s(I), Y))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)), X1))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ag(x1, x2) = log2_in_ag(x2)
U1_ag(x1, x2, x3) = U1_ag(x3)
log2_in_agg(x1, x2, x3) = log2_in_agg(x2, x3)
log2_out_agg(x1, x2, x3) = log2_out_agg(x1)
U2_agg(x1, x2, x3, x4) = U2_agg(x2, x3, x4)
half_in_aa(x1, x2) = half_in_aa
half_out_aa(x1, x2) = half_out_aa(x1, x2)
U4_aa(x1, x2, x3) = U4_aa(x3)
s(x1) = s(x1)
U3_agg(x1, x2, x3, x4) = U3_agg(x1, x4)
log2_in_ggg(x1, x2, x3) = log2_in_ggg(x1, x2, x3)
0 = 0
log2_out_ggg(x1, x2, x3) = log2_out_ggg
U2_ggg(x1, x2, x3, x4) = U2_ggg(x2, x3, x4)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U4_ga(x1, x2, x3) = U4_ga(x3)
U3_ggg(x1, x2, x3, x4) = U3_ggg(x4)
log2_out_ag(x1, x2) = log2_out_ag(x1)
LOG2_IN_GGG(x1, x2, x3) = LOG2_IN_GGG(x1, x2, x3)
U2_GGG(x1, x2, x3, x4) = U2_GGG(x2, x3, x4)

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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)), X1))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

half_in_ga(s(s(X)), s(Y)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_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)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)

The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
0 = 0
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x2)
U4_ga(x1, x2, x3) = U4_ga(x3)
LOG2_IN_GGG(x1, x2, x3) = LOG2_IN_GGG(x1, x2, x3)
U2_GGG(x1, x2, x3, x4) = U2_GGG(x2, x3, x4)

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

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(I, Y, half_in_ga(s(s(X))))
U2_GGG(I, Y, half_out_ga(X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U4_ga(half_in_ga(X))
U4_ga(half_out_ga(Y)) → half_out_ga(s(Y))
half_in_ga(0) → half_out_ga(0)
half_in_ga(s(0)) → half_out_ga(0)

The set Q consists of the following terms:

half_in_ga(x0)
U4_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 dependency pairs:

U2_GGG(I, Y, half_out_ga(X1)) → LOG2_IN_GGG(X1, s(I), Y)

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) = 0
POL(LOG2_IN_GGG(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃
POL(U2_GGG(x₁, x₂, x₃)) = 2 + x₁ + x₂ + x₃
POL(U4_ga(x₁)) = 2 + x₁
POL(half_in_ga(x₁)) = x₁
POL(half_out_ga(x₁)) = 2·x₁
POL(s(x₁)) = 1 + x₁

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ RuleRemovalProof

                          ↳ QDP

                            ↳ DependencyGraphProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(I, Y, half_in_ga(s(s(X))))

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U4_ga(half_in_ga(X))
U4_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)
U4_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.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

HALF_IN_AA(s(s(X)), s(Y)) → HALF_IN_AA(X, Y)

The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

  ↳ PrologToPiTRSProof

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

HALF_IN_AA(s(s(X)), s(Y)) → HALF_IN_AA(X, Y)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

HALF_IN_AA → HALF_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:

HALF_IN_AA → HALF_IN_AA

The TRS R consists of the following rules:none

s = HALF_IN_AA evaluates to t =HALF_IN_AA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from HALF_IN_AA to HALF_IN_AA.

We use the technique of [30]. With regard to the inferred argument filtering the predicates were used in the following modes:
log2_in: (f,b)
log2_in: (f,b,b) (b,b,b)
half_in: (f,f) (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

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:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

LOG2_IN_AG(X, Y) → U1_AG(X, Y, log2_in_agg(X, 0, Y))
LOG2_IN_AG(X, Y) → LOG2_IN_AGG(X, 0, Y)
LOG2_IN_AGG(s(s(X)), I, Y) → U2_AGG(X, I, Y, half_in_aa(s(s(X)), X1))
LOG2_IN_AGG(s(s(X)), I, Y) → HALF_IN_AA(s(s(X)), X1)
HALF_IN_AA(s(s(X)), s(Y)) → U4_AA(X, Y, half_in_aa(X, Y))
HALF_IN_AA(s(s(X)), s(Y)) → HALF_IN_AA(X, Y)
U2_AGG(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_AGG(X, I, Y, log2_in_ggg(X1, s(I), Y))
U2_AGG(X, I, Y, half_out_aa(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)
LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)), X1))
LOG2_IN_GGG(s(s(X)), I, Y) → HALF_IN_GA(s(s(X)), X1)
HALF_IN_GA(s(s(X)), s(Y)) → U4_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_GGG(X, I, Y, log2_in_ggg(X1, s(I), Y))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ag(x1, x2) = log2_in_ag(x2)
U1_ag(x1, x2, x3) = U1_ag(x2, x3)
log2_in_agg(x1, x2, x3) = log2_in_agg(x2, x3)
log2_out_agg(x1, x2, x3) = log2_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4) = U2_agg(x2, x3, x4)
half_in_aa(x1, x2) = half_in_aa
half_out_aa(x1, x2) = half_out_aa(x1, x2)
U4_aa(x1, x2, x3) = U4_aa(x3)
s(x1) = s(x1)
U3_agg(x1, x2, x3, x4) = U3_agg(x1, x2, x3, x4)
log2_in_ggg(x1, x2, x3) = log2_in_ggg(x1, x2, x3)
0 = 0
log2_out_ggg(x1, x2, x3) = log2_out_ggg(x1, x2, x3)
U2_ggg(x1, x2, x3, x4) = U2_ggg(x1, x2, x3, x4)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x1, x2)
U4_ga(x1, x2, x3) = U4_ga(x1, x3)
U3_ggg(x1, x2, x3, x4) = U3_ggg(x1, x2, x3, x4)
log2_out_ag(x1, x2) = log2_out_ag(x1, x2)
U3_AGG(x1, x2, x3, x4) = U3_AGG(x1, x2, x3, x4)
LOG2_IN_AGG(x1, x2, x3) = LOG2_IN_AGG(x2, x3)
LOG2_IN_GGG(x1, x2, x3) = LOG2_IN_GGG(x1, x2, x3)
U3_GGG(x1, x2, x3, x4) = U3_GGG(x1, x2, x3, x4)
HALF_IN_GA(x1, x2) = HALF_IN_GA(x1)
U2_AGG(x1, x2, x3, x4) = U2_AGG(x2, x3, x4)
U4_GA(x1, x2, x3) = U4_GA(x1, x3)
U2_GGG(x1, x2, x3, x4) = U2_GGG(x1, x2, x3, x4)
HALF_IN_AA(x1, x2) = HALF_IN_AA
U1_AG(x1, x2, x3) = U1_AG(x2, x3)
U4_AA(x1, x2, x3) = U4_AA(x3)
LOG2_IN_AG(x1, x2) = LOG2_IN_AG(x2)

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:

LOG2_IN_AG(X, Y) → U1_AG(X, Y, log2_in_agg(X, 0, Y))
LOG2_IN_AG(X, Y) → LOG2_IN_AGG(X, 0, Y)
LOG2_IN_AGG(s(s(X)), I, Y) → U2_AGG(X, I, Y, half_in_aa(s(s(X)), X1))
LOG2_IN_AGG(s(s(X)), I, Y) → HALF_IN_AA(s(s(X)), X1)
HALF_IN_AA(s(s(X)), s(Y)) → U4_AA(X, Y, half_in_aa(X, Y))
HALF_IN_AA(s(s(X)), s(Y)) → HALF_IN_AA(X, Y)
U2_AGG(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_AGG(X, I, Y, log2_in_ggg(X1, s(I), Y))
U2_AGG(X, I, Y, half_out_aa(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)
LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)), X1))
LOG2_IN_GGG(s(s(X)), I, Y) → HALF_IN_GA(s(s(X)), X1)
HALF_IN_GA(s(s(X)), s(Y)) → U4_GA(X, Y, half_in_ga(X, Y))
HALF_IN_GA(s(s(X)), s(Y)) → HALF_IN_GA(X, Y)
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_GGG(X, I, Y, log2_in_ggg(X1, s(I), Y))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

              ↳ 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:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

              ↳ 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)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ PiDP

              ↳ PiDP

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

HALF_IN_GA(s(s(X))) → HALF_IN_GA(X)

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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)), X1))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

The argument filtering Pi contains the following mapping:
log2_in_ag(x1, x2) = log2_in_ag(x2)
U1_ag(x1, x2, x3) = U1_ag(x2, x3)
log2_in_agg(x1, x2, x3) = log2_in_agg(x2, x3)
log2_out_agg(x1, x2, x3) = log2_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4) = U2_agg(x2, x3, x4)
half_in_aa(x1, x2) = half_in_aa
half_out_aa(x1, x2) = half_out_aa(x1, x2)
U4_aa(x1, x2, x3) = U4_aa(x3)
s(x1) = s(x1)
U3_agg(x1, x2, x3, x4) = U3_agg(x1, x2, x3, x4)
log2_in_ggg(x1, x2, x3) = log2_in_ggg(x1, x2, x3)
0 = 0
log2_out_ggg(x1, x2, x3) = log2_out_ggg(x1, x2, x3)
U2_ggg(x1, x2, x3, x4) = U2_ggg(x1, x2, x3, x4)
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x1, x2)
U4_ga(x1, x2, x3) = U4_ga(x1, x3)
U3_ggg(x1, x2, x3, x4) = U3_ggg(x1, x2, x3, x4)
log2_out_ag(x1, x2) = log2_out_ag(x1, x2)
LOG2_IN_GGG(x1, x2, x3) = LOG2_IN_GGG(x1, x2, x3)
U2_GGG(x1, x2, x3, x4) = U2_GGG(x1, x2, x3, x4)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

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

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)), X1))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

half_in_ga(s(s(X)), s(Y)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_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)
half_in_ga(s(0), 0) → half_out_ga(s(0), 0)

The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
0 = 0
half_in_ga(x1, x2) = half_in_ga(x1)
half_out_ga(x1, x2) = half_out_ga(x1, x2)
U4_ga(x1, x2, x3) = U4_ga(x1, x3)
LOG2_IN_GGG(x1, x2, x3) = LOG2_IN_GGG(x1, x2, x3)
U2_GGG(x1, x2, x3, x4) = U2_GGG(x1, x2, x3, x4)

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

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Rewriting

              ↳ PiDP

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

U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)
LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X))))

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U4_ga(X, half_in_ga(X))
U4_ga(X, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
half_in_ga(0) → half_out_ga(0, 0)
half_in_ga(s(0)) → half_out_ga(s(0), 0)

The set Q consists of the following terms:

half_in_ga(x0)
U4_ga(x0, x1)

We have to consider all (P,Q,R)-chains.
By rewriting [15] the rule LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, half_in_ga(s(s(X)))) at position [3] we obtained the following new rules:

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, U4_ga(X, half_in_ga(X)))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ QDPOrderProof

              ↳ PiDP

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

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, U4_ga(X, half_in_ga(X)))
U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U4_ga(X, half_in_ga(X))
U4_ga(X, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
half_in_ga(0) → half_out_ga(0, 0)
half_in_ga(s(0)) → half_out_ga(s(0), 0)

The set Q consists of the following terms:

half_in_ga(x0)
U4_ga(x0, x1)

We have to consider all (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

U2_GGG(X, I, Y, half_out_ga(s(s(X)), X1)) → LOG2_IN_GGG(X1, s(I), Y)

The remaining pairs can at least be oriented weakly.

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, U4_ga(X, half_in_ga(X)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(LOG2_IN_GGG(x₁, x₂, x₃)) = x₁
POL(U2_GGG(x₁, x₂, x₃, x₄)) = 1 + x₄
POL(U4_ga(x₁, x₂)) = 1 + x₂
POL(half_in_ga(x₁)) = x₁
POL(half_out_ga(x₁, x₂)) = x₂
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

U4_ga(X, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
half_in_ga(0) → half_out_ga(0, 0)
half_in_ga(s(s(X))) → U4_ga(X, half_in_ga(X))
half_in_ga(s(0)) → half_out_ga(s(0), 0)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Rewriting

                          ↳ QDP

                            ↳ QDPOrderProof

                              ↳ QDP

                                ↳ DependencyGraphProof

              ↳ PiDP

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

LOG2_IN_GGG(s(s(X)), I, Y) → U2_GGG(X, I, Y, U4_ga(X, half_in_ga(X)))

The TRS R consists of the following rules:

half_in_ga(s(s(X))) → U4_ga(X, half_in_ga(X))
U4_ga(X, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
half_in_ga(0) → half_out_ga(0, 0)
half_in_ga(s(0)) → half_out_ga(s(0), 0)

The set Q consists of the following terms:

half_in_ga(x0)
U4_ga(x0, x1)

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.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

HALF_IN_AA(s(s(X)), s(Y)) → HALF_IN_AA(X, Y)

The TRS R consists of the following rules:

log2_in_ag(X, Y) → U1_ag(X, Y, log2_in_agg(X, 0, Y))
log2_in_agg(0, I, I) → log2_out_agg(0, I, I)
log2_in_agg(s(0), I, I) → log2_out_agg(s(0), I, I)
log2_in_agg(s(s(X)), I, Y) → U2_agg(X, I, Y, half_in_aa(s(s(X)), X1))
half_in_aa(0, 0) → half_out_aa(0, 0)
half_in_aa(s(0), 0) → half_out_aa(s(0), 0)
half_in_aa(s(s(X)), s(Y)) → U4_aa(X, Y, half_in_aa(X, Y))
U4_aa(X, Y, half_out_aa(X, Y)) → half_out_aa(s(s(X)), s(Y))
U2_agg(X, I, Y, half_out_aa(s(s(X)), X1)) → U3_agg(X, I, Y, log2_in_ggg(X1, s(I), Y))
log2_in_ggg(0, I, I) → log2_out_ggg(0, I, I)
log2_in_ggg(s(0), I, I) → log2_out_ggg(s(0), I, I)
log2_in_ggg(s(s(X)), I, Y) → U2_ggg(X, I, Y, half_in_ga(s(s(X)), X1))
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)) → U4_ga(X, Y, half_in_ga(X, Y))
U4_ga(X, Y, half_out_ga(X, Y)) → half_out_ga(s(s(X)), s(Y))
U2_ggg(X, I, Y, half_out_ga(s(s(X)), X1)) → U3_ggg(X, I, Y, log2_in_ggg(X1, s(I), Y))
U3_ggg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_ggg(s(s(X)), I, Y)
U3_agg(X, I, Y, log2_out_ggg(X1, s(I), Y)) → log2_out_agg(s(s(X)), I, Y)
U1_ag(X, Y, log2_out_agg(X, 0, Y)) → log2_out_ag(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

HALF_IN_AA(s(s(X)), s(Y)) → HALF_IN_AA(X, Y)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

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

HALF_IN_AA → HALF_IN_AA

The TRS R consists of the following rules:none

s = HALF_IN_AA evaluates to t =HALF_IN_AA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [ ]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from HALF_IN_AA to HALF_IN_AA.