Left Termination proof of /tmp/tmp30APd-/in-fb.pl

Left Termination of the query pattern in_in_2(a, g) w.r.t. the given Prolog program could not be shown:

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

Clauses:

in(X, tree(X, X1, X2)).
in(X, tree(Y, Left, X1)) :- ','(less(X, Y), in(X, Left)).
in(X, tree(Y, X1, Right)) :- ','(less(Y, X), in(X, Right)).
less(0, s(X)).
less(s(X), s(Y)) :- less(X, Y).

Queries:

in(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:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

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:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2) = in_in(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x4, x5)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
U4(x1, x2, x3, x4, x5) = U4(x5)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
in_out(x1, x2) = in_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:

IN_IN(X, tree(Y, X1, Right)) → U3¹(X, Y, X1, Right, less_in(Y, X))
IN_IN(X, tree(Y, X1, Right)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U5¹(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U3¹(X, Y, X1, Right, less_out(Y, X)) → U4¹(X, Y, X1, Right, in_in(X, Right))
U3¹(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
IN_IN(X, tree(Y, Left, X1)) → U1¹(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, Left, X1)) → LESS_IN(X, Y)
U1¹(X, Y, Left, X1, less_out(X, Y)) → U2¹(X, Y, Left, X1, in_in(X, Left))
U1¹(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2) = in_in(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x4, x5)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
U4(x1, x2, x3, x4, x5) = U4(x5)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
in_out(x1, x2) = in_out(x1)
IN_IN(x1, x2) = IN_IN(x2)
U5¹(x1, x2, x3) = U5¹(x3)
LESS_IN(x1, x2) = LESS_IN
U4¹(x1, x2, x3, x4, x5) = U4¹(x5)
U3¹(x1, x2, x3, x4, x5) = U3¹(x4, x5)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x5)

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:

IN_IN(X, tree(Y, X1, Right)) → U3¹(X, Y, X1, Right, less_in(Y, X))
IN_IN(X, tree(Y, X1, Right)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U5¹(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U3¹(X, Y, X1, Right, less_out(Y, X)) → U4¹(X, Y, X1, Right, in_in(X, Right))
U3¹(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
IN_IN(X, tree(Y, Left, X1)) → U1¹(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, Left, X1)) → LESS_IN(X, Y)
U1¹(X, Y, Left, X1, less_out(X, Y)) → U2¹(X, Y, Left, X1, in_in(X, Left))
U1¹(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2) = in_in(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x4, x5)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
U4(x1, x2, x3, x4, x5) = U4(x5)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
in_out(x1, x2) = in_out(x1)
IN_IN(x1, x2) = IN_IN(x2)
U5¹(x1, x2, x3) = U5¹(x3)
LESS_IN(x1, x2) = LESS_IN
U4¹(x1, x2, x3, x4, x5) = U4¹(x5)
U3¹(x1, x2, x3, x4, x5) = U3¹(x4, x5)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x5)

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:

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2) = in_in(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x4, x5)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
U4(x1, x2, x3, x4, x5) = U4(x5)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
in_out(x1, x2) = in_out(x1)
LESS_IN(x1, x2) = LESS_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:

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

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

LESS_IN → LESS_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:

LESS_IN → LESS_IN

The TRS R consists of the following rules:none

s = LESS_IN evaluates to t =LESS_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 LESS_IN to LESS_IN.

↳ Prolog

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

  ↳ PrologToPiTRSProof

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

IN_IN(X, tree(Y, Left, X1)) → U1¹(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, X1, Right)) → U3¹(X, Y, X1, Right, less_in(Y, X))
U3¹(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
U1¹(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2) = in_in(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x4, x5)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
U4(x1, x2, x3, x4, x5) = U4(x5)
U1(x1, x2, x3, x4, x5) = U1(x3, x5)
U2(x1, x2, x3, x4, x5) = U2(x1, x5)
in_out(x1, x2) = in_out(x1)
IN_IN(x1, x2) = IN_IN(x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x4, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x5)

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:

IN_IN(X, tree(Y, Left, X1)) → U1¹(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, X1, Right)) → U3¹(X, Y, X1, Right, less_in(Y, X))
U3¹(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
U1¹(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
tree(x1, x2, x3) = tree(x1, x2, x3)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
IN_IN(x1, x2) = IN_IN(x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x4, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x3, x5)

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

                        ↳ QDPSizeChangeProof

  ↳ PrologToPiTRSProof

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

U1¹(Left, less_out(X)) → IN_IN(Left)
U3¹(Right, less_out(Y)) → IN_IN(Right)
IN_IN(tree(Y, Left, X1)) → U1¹(Left, less_in)
IN_IN(tree(Y, X1, Right)) → U3¹(Right, less_in)

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U5(x0)

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:

U1¹(Left, less_out(X)) → IN_IN(Left)
The graph contains the following edges 1 >= 1
U3¹(Right, less_out(Y)) → IN_IN(Right)
The graph contains the following edges 1 >= 1
IN_IN(tree(Y, X1, Right)) → U3¹(Right, less_in)
The graph contains the following edges 1 > 1
IN_IN(tree(Y, Left, X1)) → U1¹(Left, less_in)
The graph contains the following edges 1 > 1

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:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

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:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

IN_IN(X, tree(Y, X1, Right)) → U3¹(X, Y, X1, Right, less_in(Y, X))
IN_IN(X, tree(Y, X1, Right)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U5¹(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U3¹(X, Y, X1, Right, less_out(Y, X)) → U4¹(X, Y, X1, Right, in_in(X, Right))
U3¹(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
IN_IN(X, tree(Y, Left, X1)) → U1¹(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, Left, X1)) → LESS_IN(X, Y)
U1¹(X, Y, Left, X1, less_out(X, Y)) → U2¹(X, Y, Left, X1, in_in(X, Left))
U1¹(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2) = in_in(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x2, x3, x4, x5)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
U4(x1, x2, x3, x4, x5) = U4(x2, x3, x4, x5)
U1(x1, x2, x3, x4, x5) = U1(x2, x3, x4, x5)
U2(x1, x2, x3, x4, x5) = U2(x1, x2, x3, x4, x5)
in_out(x1, x2) = in_out(x1, x2)
IN_IN(x1, x2) = IN_IN(x2)
U5¹(x1, x2, x3) = U5¹(x3)
LESS_IN(x1, x2) = LESS_IN
U4¹(x1, x2, x3, x4, x5) = U4¹(x2, x3, x4, x5)
U3¹(x1, x2, x3, x4, x5) = U3¹(x2, x3, x4, x5)
U2¹(x1, x2, x3, x4, x5) = U2¹(x1, x2, x3, x4, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x3, x4, x5)

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:

IN_IN(X, tree(Y, X1, Right)) → U3¹(X, Y, X1, Right, less_in(Y, X))
IN_IN(X, tree(Y, X1, Right)) → LESS_IN(Y, X)
LESS_IN(s(X), s(Y)) → U5¹(X, Y, less_in(X, Y))
LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)
U3¹(X, Y, X1, Right, less_out(Y, X)) → U4¹(X, Y, X1, Right, in_in(X, Right))
U3¹(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
IN_IN(X, tree(Y, Left, X1)) → U1¹(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, Left, X1)) → LESS_IN(X, Y)
U1¹(X, Y, Left, X1, less_out(X, Y)) → U2¹(X, Y, Left, X1, in_in(X, Left))
U1¹(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

                ↳ UsableRulesProof

              ↳ PiDP

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

LESS_IN(s(X), s(Y)) → LESS_IN(X, Y)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2) = in_in(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x2, x3, x4, x5)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
U4(x1, x2, x3, x4, x5) = U4(x2, x3, x4, x5)
U1(x1, x2, x3, x4, x5) = U1(x2, x3, x4, x5)
U2(x1, x2, x3, x4, x5) = U2(x1, x2, x3, x4, x5)
in_out(x1, x2) = in_out(x1, x2)
LESS_IN(x1, x2) = LESS_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:

LESS_IN(s(X), s(Y)) → LESS_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:

LESS_IN → LESS_IN

The TRS R consists of the following rules:none

s = LESS_IN evaluates to t =LESS_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 LESS_IN to LESS_IN.

↳ Prolog

  ↳ PrologToPiTRSProof

  ↳ PrologToPiTRSProof

    ↳ PiTRS

      ↳ DependencyPairsProof

        ↳ PiDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ PiDP

              ↳ PiDP

                ↳ UsableRulesProof

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

IN_IN(X, tree(Y, Left, X1)) → U1¹(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, X1, Right)) → U3¹(X, Y, X1, Right, less_in(Y, X))
U3¹(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
U1¹(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

in_in(X, tree(Y, X1, Right)) → U3(X, Y, X1, Right, less_in(Y, X))
less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))
U3(X, Y, X1, Right, less_out(Y, X)) → U4(X, Y, X1, Right, in_in(X, Right))
in_in(X, tree(Y, Left, X1)) → U1(X, Y, Left, X1, less_in(X, Y))
U1(X, Y, Left, X1, less_out(X, Y)) → U2(X, Y, Left, X1, in_in(X, Left))
in_in(X, tree(X, X1, X2)) → in_out(X, tree(X, X1, X2))
U2(X, Y, Left, X1, in_out(X, Left)) → in_out(X, tree(Y, Left, X1))
U4(X, Y, X1, Right, in_out(X, Right)) → in_out(X, tree(Y, X1, Right))

The argument filtering Pi contains the following mapping:
in_in(x1, x2) = in_in(x2)
tree(x1, x2, x3) = tree(x1, x2, x3)
U3(x1, x2, x3, x4, x5) = U3(x2, x3, x4, x5)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
U4(x1, x2, x3, x4, x5) = U4(x2, x3, x4, x5)
U1(x1, x2, x3, x4, x5) = U1(x2, x3, x4, x5)
U2(x1, x2, x3, x4, x5) = U2(x1, x2, x3, x4, x5)
in_out(x1, x2) = in_out(x1, x2)
IN_IN(x1, x2) = IN_IN(x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x2, x3, x4, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x3, x4, x5)

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:

IN_IN(X, tree(Y, Left, X1)) → U1¹(X, Y, Left, X1, less_in(X, Y))
IN_IN(X, tree(Y, X1, Right)) → U3¹(X, Y, X1, Right, less_in(Y, X))
U3¹(X, Y, X1, Right, less_out(Y, X)) → IN_IN(X, Right)
U1¹(X, Y, Left, X1, less_out(X, Y)) → IN_IN(X, Left)

The TRS R consists of the following rules:

less_in(s(X), s(Y)) → U5(X, Y, less_in(X, Y))
less_in(0, s(X)) → less_out(0, s(X))
U5(X, Y, less_out(X, Y)) → less_out(s(X), s(Y))

The argument filtering Pi contains the following mapping:
tree(x1, x2, x3) = tree(x1, x2, x3)
less_in(x1, x2) = less_in
s(x1) = s(x1)
U5(x1, x2, x3) = U5(x3)
0 = 0
less_out(x1, x2) = less_out(x1)
IN_IN(x1, x2) = IN_IN(x2)
U3¹(x1, x2, x3, x4, x5) = U3¹(x2, x3, x4, x5)
U1¹(x1, x2, x3, x4, x5) = U1¹(x2, x3, x4, x5)

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

                        ↳ QDPSizeChangeProof

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

U1¹(Y, Left, X1, less_out(X)) → IN_IN(Left)
IN_IN(tree(Y, X1, Right)) → U3¹(Y, X1, Right, less_in)
IN_IN(tree(Y, Left, X1)) → U1¹(Y, Left, X1, less_in)
U3¹(Y, X1, Right, less_out(Y)) → IN_IN(Right)

The TRS R consists of the following rules:

less_in → U5(less_in)
less_in → less_out(0)
U5(less_out(X)) → less_out(s(X))

The set Q consists of the following terms:

less_in
U5(x0)

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:

U1¹(Y, Left, X1, less_out(X)) → IN_IN(Left)
The graph contains the following edges 2 >= 1
U3¹(Y, X1, Right, less_out(Y)) → IN_IN(Right)
The graph contains the following edges 3 >= 1
IN_IN(tree(Y, X1, Right)) → U3¹(Y, X1, Right, less_in)
The graph contains the following edges 1 > 1, 1 > 2, 1 > 3
IN_IN(tree(Y, Left, X1)) → U1¹(Y, Left, X1, less_in)
The graph contains the following edges 1 > 1, 1 > 2, 1 > 3