Left Termination proof of /tmp/tmpOIH

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

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(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(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x4)
log2_out(x1, x2, x3) = log2_out(x1)
log2_out(x1, x2) = log2_out(x1)

Using Dependency Pairs [1,30] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

LOG2_IN(X, Y) → U1¹(X, Y, log2_in(X, 0, Y))
LOG2_IN(X, Y) → LOG2_IN(X, 0, Y)
LOG2_IN(s(s(X)), I, Y) → U2¹(X, I, Y, half_in(s(s(X)), X1))
LOG2_IN(s(s(X)), I, Y) → HALF_IN(s(s(X)), X1)
HALF_IN(s(s(X)), s(Y)) → U4¹(X, Y, half_in(X, Y))
HALF_IN(s(s(X)), s(Y)) → HALF_IN(X, Y)
U2¹(X, I, Y, half_out(s(s(X)), X1)) → U3¹(X, I, Y, log2_in(X1, s(I), Y))
U2¹(X, I, Y, half_out(s(s(X)), X1)) → LOG2_IN(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x4)
log2_out(x1, x2, x3) = log2_out(x1)
log2_out(x1, x2) = log2_out(x1)
U3¹(x1, x2, x3, x4) = U3¹(x1, x4)
U4¹(x1, x2, x3) = U4¹(x3)
HALF_IN(x1, x2) = HALF_IN
LOG2_IN(x1, x2) = LOG2_IN(x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
LOG2_IN(x1, x2, x3) = LOG2_IN(x2, 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:

LOG2_IN(X, Y) → U1¹(X, Y, log2_in(X, 0, Y))
LOG2_IN(X, Y) → LOG2_IN(X, 0, Y)
LOG2_IN(s(s(X)), I, Y) → U2¹(X, I, Y, half_in(s(s(X)), X1))
LOG2_IN(s(s(X)), I, Y) → HALF_IN(s(s(X)), X1)
HALF_IN(s(s(X)), s(Y)) → U4¹(X, Y, half_in(X, Y))
HALF_IN(s(s(X)), s(Y)) → HALF_IN(X, Y)
U2¹(X, I, Y, half_out(s(s(X)), X1)) → U3¹(X, I, Y, log2_in(X1, s(I), Y))
U2¹(X, I, Y, half_out(s(s(X)), X1)) → LOG2_IN(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x4)
log2_out(x1, x2, x3) = log2_out(x1)
log2_out(x1, x2) = log2_out(x1)
U3¹(x1, x2, x3, x4) = U3¹(x1, x4)
U4¹(x1, x2, x3) = U4¹(x3)
HALF_IN(x1, x2) = HALF_IN
LOG2_IN(x1, x2) = LOG2_IN(x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
LOG2_IN(x1, x2, x3) = LOG2_IN(x2, x3)
U1¹(x1, x2, x3) = U1¹(x3)

We have to consider all (P,R,Pi)-chains
The approximation of the Dependency Graph [30] contains 2 SCCs with 5 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x4)
log2_out(x1, x2, x3) = log2_out(x1)
log2_out(x1, x2) = log2_out(x1)
HALF_IN(x1, x2) = HALF_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

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

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

                        ↳ NonTerminationProof

              ↳ PiDP

  ↳ PrologToPiTRSProof

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

HALF_IN → HALF_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:

HALF_IN → HALF_IN

The TRS R consists of the following rules:none

s = HALF_IN evaluates to t =HALF_IN

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 to HALF_IN.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

LOG2_IN(s(s(X)), I, Y) → U2¹(X, I, Y, half_in(s(s(X)), X1))
U2¹(X, I, Y, half_out(s(s(X)), X1)) → LOG2_IN(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x4)
log2_out(x1, x2, x3) = log2_out(x1)
log2_out(x1, x2) = log2_out(x1)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
LOG2_IN(x1, x2, x3) = LOG2_IN(x2, 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

  ↳ PrologToPiTRSProof

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

LOG2_IN(s(s(X)), I, Y) → U2¹(X, I, Y, half_in(s(s(X)), X1))
U2¹(X, I, Y, half_out(s(s(X)), X1)) → LOG2_IN(X1, s(I), Y)

The TRS R consists of the following rules:

half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)

The argument filtering Pi contains the following mapping:
0 = 0
s(x1) = s(x1)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
LOG2_IN(x1, x2, x3) = LOG2_IN(x2, 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

                        ↳ Narrowing

  ↳ PrologToPiTRSProof

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

LOG2_IN(I, Y) → U2¹(I, Y, half_in)
U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule LOG2_IN(I, Y) → U2¹(I, Y, half_in) at position [2] we obtained the following new rules:

LOG2_IN(y0, y1) → U2¹(y0, y1, half_out(s(0), 0))
LOG2_IN(y0, y1) → U2¹(y0, y1, half_out(0, 0))
LOG2_IN(y0, y1) → U2¹(y0, y1, U4(half_in))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

  ↳ PrologToPiTRSProof

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

LOG2_IN(y0, y1) → U2¹(y0, y1, half_out(s(0), 0))
LOG2_IN(y0, y1) → U2¹(y0, y1, half_out(0, 0))
LOG2_IN(y0, y1) → U2¹(y0, y1, U4(half_in))
U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

  ↳ PrologToPiTRSProof

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

U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y)
LOG2_IN(y0, y1) → U2¹(y0, y1, U4(half_in))

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule LOG2_IN(y0, y1) → U2¹(y0, y1, U4(half_in)) we obtained the following new rules:

LOG2_IN(s(z0), z1) → U2¹(s(z0), z1, U4(half_in))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

  ↳ PrologToPiTRSProof

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

LOG2_IN(s(z0), z1) → U2¹(s(z0), z1, U4(half_in))
U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y) we obtained the following new rules:

U2¹(s(z0), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(z0)), z1)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

  ↳ PrologToPiTRSProof

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

U2¹(s(z0), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(z0)), z1)
LOG2_IN(s(z0), z1) → U2¹(s(z0), z1, U4(half_in))

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule LOG2_IN(s(z0), z1) → U2¹(s(z0), z1, U4(half_in)) we obtained the following new rules:

LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Instantiation

  ↳ PrologToPiTRSProof

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

LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))
U2¹(s(z0), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(z0)), z1)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U2¹(s(z0), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(z0)), z1) we obtained the following new rules:

U2¹(s(s(z0)), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(s(z0))), z1)

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ NonTerminationProof

  ↳ PrologToPiTRSProof

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

LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))
U2¹(s(s(z0)), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(s(z0))), z1)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

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 narrowing to the left:

The TRS P consists of the following rules:

LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))
U2¹(s(s(z0)), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(s(z0))), z1)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

s = U2¹(s(s(z0')), z1', U4(half_in)) evaluates to t =U2¹(s(s(s(z0'))), z1', U4(half_in))

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

Semiunifier: [ ]
Matcher: [z0' / s(z0')]

Rewriting sequence

U2¹(s(s(z0')), z1', U4(half_in)) → U2¹(s(s(z0')), z1', U4(half_out(s(0), 0)))
with rule half_in → half_out(s(0), 0) at position [2,0] and matcher [ ]

U2¹(s(s(z0')), z1', U4(half_out(s(0), 0))) → U2¹(s(s(z0')), z1', half_out(s(s(s(0))), s(0)))
with rule U4(half_out(X, Y)) → half_out(s(s(X)), s(Y)) at position [2] and matcher [X / s(0), Y / 0]

U2¹(s(s(z0')), z1', half_out(s(s(s(0))), s(0))) → LOG2_IN(s(s(s(z0'))), z1')
with rule U2¹(s(s(z0'')), z1'', half_out(s(s(x2)), x3)) → LOG2_IN(s(s(s(z0''))), z1'') at position [] and matcher [x3 / s(0), x2 / s(0), z1'' / z1', z0'' / z0']

LOG2_IN(s(s(s(z0'))), z1') → U2¹(s(s(s(z0'))), z1', U4(half_in))
with rule LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence

All these steps are and every following step will be a correct step w.r.t to Q.

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:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(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(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x2, x3, x4)
log2_out(x1, x2, x3) = log2_out(x1, x2, x3)
log2_out(x1, x2) = log2_out(x1, 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:

LOG2_IN(X, Y) → U1¹(X, Y, log2_in(X, 0, Y))
LOG2_IN(X, Y) → LOG2_IN(X, 0, Y)
LOG2_IN(s(s(X)), I, Y) → U2¹(X, I, Y, half_in(s(s(X)), X1))
LOG2_IN(s(s(X)), I, Y) → HALF_IN(s(s(X)), X1)
HALF_IN(s(s(X)), s(Y)) → U4¹(X, Y, half_in(X, Y))
HALF_IN(s(s(X)), s(Y)) → HALF_IN(X, Y)
U2¹(X, I, Y, half_out(s(s(X)), X1)) → U3¹(X, I, Y, log2_in(X1, s(I), Y))
U2¹(X, I, Y, half_out(s(s(X)), X1)) → LOG2_IN(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x2, x3, x4)
log2_out(x1, x2, x3) = log2_out(x1, x2, x3)
log2_out(x1, x2) = log2_out(x1, x2)
U3¹(x1, x2, x3, x4) = U3¹(x1, x2, x3, x4)
U4¹(x1, x2, x3) = U4¹(x3)
HALF_IN(x1, x2) = HALF_IN
LOG2_IN(x1, x2) = LOG2_IN(x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
LOG2_IN(x1, x2, x3) = LOG2_IN(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:

LOG2_IN(X, Y) → U1¹(X, Y, log2_in(X, 0, Y))
LOG2_IN(X, Y) → LOG2_IN(X, 0, Y)
LOG2_IN(s(s(X)), I, Y) → U2¹(X, I, Y, half_in(s(s(X)), X1))
LOG2_IN(s(s(X)), I, Y) → HALF_IN(s(s(X)), X1)
HALF_IN(s(s(X)), s(Y)) → U4¹(X, Y, half_in(X, Y))
HALF_IN(s(s(X)), s(Y)) → HALF_IN(X, Y)
U2¹(X, I, Y, half_out(s(s(X)), X1)) → U3¹(X, I, Y, log2_in(X1, s(I), Y))
U2¹(X, I, Y, half_out(s(s(X)), X1)) → LOG2_IN(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x2, x3, x4)
log2_out(x1, x2, x3) = log2_out(x1, x2, x3)
log2_out(x1, x2) = log2_out(x1, x2)
U3¹(x1, x2, x3, x4) = U3¹(x1, x2, x3, x4)
U4¹(x1, x2, x3) = U4¹(x3)
HALF_IN(x1, x2) = HALF_IN
LOG2_IN(x1, x2) = LOG2_IN(x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
LOG2_IN(x1, x2, x3) = LOG2_IN(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 2 SCCs with 5 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ 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:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x2, x3, x4)
log2_out(x1, x2, x3) = log2_out(x1, x2, x3)
log2_out(x1, x2) = log2_out(x1, x2)
HALF_IN(x1, x2) = HALF_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

            ↳ 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)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ NonTerminationProof

              ↳ PiDP

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

HALF_IN → HALF_IN

The TRS R consists of the following rules:none

s = HALF_IN evaluates to t =HALF_IN

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 to HALF_IN.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

LOG2_IN(s(s(X)), I, Y) → U2¹(X, I, Y, half_in(s(s(X)), X1))
U2¹(X, I, Y, half_out(s(s(X)), X1)) → LOG2_IN(X1, s(I), Y)

The TRS R consists of the following rules:

log2_in(X, Y) → U1(X, Y, log2_in(X, 0, Y))
log2_in(s(s(X)), I, Y) → U2(X, I, Y, half_in(s(s(X)), X1))
half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
U2(X, I, Y, half_out(s(s(X)), X1)) → U3(X, I, Y, log2_in(X1, s(I), Y))
log2_in(s(0), I, I) → log2_out(s(0), I, I)
log2_in(0, I, I) → log2_out(0, I, I)
U3(X, I, Y, log2_out(X1, s(I), Y)) → log2_out(s(s(X)), I, Y)
U1(X, Y, log2_out(X, 0, Y)) → log2_out(X, Y)

The argument filtering Pi contains the following mapping:
log2_in(x1, x2) = log2_in(x2)
U1(x1, x2, x3) = U1(x2, x3)
log2_in(x1, x2, x3) = log2_in(x2, x3)
0 = 0
s(x1) = s(x1)
U2(x1, x2, x3, x4) = U2(x2, x3, x4)
half_in(x1, x2) = half_in
U4(x1, x2, x3) = U4(x3)
half_out(x1, x2) = half_out(x1, x2)
U3(x1, x2, x3, x4) = U3(x1, x2, x3, x4)
log2_out(x1, x2, x3) = log2_out(x1, x2, x3)
log2_out(x1, x2) = log2_out(x1, x2)
U2¹(x1, x2, x3, x4) = U2¹(x2, x3, x4)
LOG2_IN(x1, x2, x3) = LOG2_IN(x2, 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

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

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

LOG2_IN(s(s(X)), I, Y) → U2¹(X, I, Y, half_in(s(s(X)), X1))
U2¹(X, I, Y, half_out(s(s(X)), X1)) → LOG2_IN(X1, s(I), Y)

The TRS R consists of the following rules:

half_in(s(s(X)), s(Y)) → U4(X, Y, half_in(X, Y))
U4(X, Y, half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in(s(0), 0) → half_out(s(0), 0)
half_in(0, 0) → half_out(0, 0)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

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

LOG2_IN(I, Y) → U2¹(I, Y, half_in)
U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By narrowing [15] the rule LOG2_IN(I, Y) → U2¹(I, Y, half_in) at position [2] we obtained the following new rules:

LOG2_IN(y0, y1) → U2¹(y0, y1, half_out(s(0), 0))
LOG2_IN(y0, y1) → U2¹(y0, y1, half_out(0, 0))
LOG2_IN(y0, y1) → U2¹(y0, y1, U4(half_in))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

LOG2_IN(y0, y1) → U2¹(y0, y1, half_out(s(0), 0))
LOG2_IN(y0, y1) → U2¹(y0, y1, half_out(0, 0))
LOG2_IN(y0, y1) → U2¹(y0, y1, U4(half_in))
U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

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

U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y)
LOG2_IN(y0, y1) → U2¹(y0, y1, U4(half_in))

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule LOG2_IN(y0, y1) → U2¹(y0, y1, U4(half_in)) we obtained the following new rules:

LOG2_IN(s(z0), z1) → U2¹(s(z0), z1, U4(half_in))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

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

LOG2_IN(s(z0), z1) → U2¹(s(z0), z1, U4(half_in))
U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U2¹(I, Y, half_out(s(s(X)), X1)) → LOG2_IN(s(I), Y) we obtained the following new rules:

U2¹(s(z0), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(z0)), z1)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

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

U2¹(s(z0), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(z0)), z1)
LOG2_IN(s(z0), z1) → U2¹(s(z0), z1, U4(half_in))

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule LOG2_IN(s(z0), z1) → U2¹(s(z0), z1, U4(half_in)) we obtained the following new rules:

LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Instantiation

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

LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))
U2¹(s(z0), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(z0)), z1)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

We have to consider all (P,Q,R)-chains.
By instantiating [15] the rule U2¹(s(z0), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(z0)), z1) we obtained the following new rules:

U2¹(s(s(z0)), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(s(z0))), z1)

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

                  ↳ PiDP

                    ↳ PiDPToQDPProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ Instantiation

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Instantiation

                                              ↳ QDP

                                                ↳ NonTerminationProof

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

LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))
U2¹(s(s(z0)), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(s(z0))), z1)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

The set Q consists of the following terms:

half_in
U4(x0)

LOG2_IN(s(s(z0)), z1) → U2¹(s(s(z0)), z1, U4(half_in))
U2¹(s(s(z0)), z1, half_out(s(s(x2)), x3)) → LOG2_IN(s(s(s(z0))), z1)

The TRS R consists of the following rules:

half_in → U4(half_in)
U4(half_out(X, Y)) → half_out(s(s(X)), s(Y))
half_in → half_out(s(0), 0)
half_in → half_out(0, 0)

s = U2¹(s(s(z0')), z1', U4(half_in)) evaluates to t =U2¹(s(s(s(z0'))), z1', U4(half_in))

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

Semiunifier: [ ]
Matcher: [z0' / s(z0')]