Termination w.r.t. Q proof of /tmp/tmp4X9Kmq/ternary-hard.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

*¹(x, +(y, z)) → *¹(x, y)
+¹(0(x), 0(y)) → 0¹(+(x, y))
*¹(+(x, y), z) → *¹(x, z)
*¹(j(x), y) → *¹(x, y)
+¹(j(x), 0(y)) → +¹(x, y)
+¹(0(x), j(y)) → +¹(x, y)
*¹(0(x), y) → 0¹(*(x, y))
+¹(1(x), j(y)) → +¹(x, y)
+¹(j(x), 1(y)) → +¹(x, y)
+¹(1(x), 1(y)) → +¹(x, y)
OPP(j(x)) → OPP(x)
*¹(1(x), y) → +¹(0(*(x, y)), y)
+¹(j(x), j(y)) → +¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
*¹(1(x), y) → 0¹(*(x, y))
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
*¹(+(x, y), z) → *¹(y, z)
-¹(x, y) → +¹(x, opp(y))
+¹(j(x), 1(y)) → 0¹(+(x, y))
+¹(1(x), j(y)) → 0¹(+(x, y))
*¹(0(x), y) → *¹(x, y)
OPP(0(x)) → 0¹(opp(x))
OPP(0(x)) → OPP(x)
+¹(+(x, y), z) → +¹(y, z)
OPP(1(x)) → OPP(x)
*¹(j(x), y) → -¹(0(*(x, y)), y)
*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))
+¹(0(x), 0(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(0(x), 1(y)) → +¹(x, y)
-¹(x, y) → OPP(y)
*¹(*(x, y), z) → *¹(y, z)
*¹(+(x, y), z) → +¹(*(x, z), *(y, z))
+¹(+(x, y), z) → +¹(x, +(y, z))
*¹(*(x, y), z) → *¹(x, *(y, z))
+¹(j(x), j(y)) → +¹(+(x, y), j(#))
*¹(j(x), y) → 0¹(*(x, y))
*¹(1(x), y) → *¹(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

*¹(x, +(y, z)) → *¹(x, y)
+¹(0(x), 0(y)) → 0¹(+(x, y))
*¹(+(x, y), z) → *¹(x, z)
*¹(j(x), y) → *¹(x, y)
+¹(j(x), 0(y)) → +¹(x, y)
+¹(0(x), j(y)) → +¹(x, y)
*¹(0(x), y) → 0¹(*(x, y))
+¹(1(x), j(y)) → +¹(x, y)
+¹(j(x), 1(y)) → +¹(x, y)
+¹(1(x), 1(y)) → +¹(x, y)
OPP(j(x)) → OPP(x)
*¹(1(x), y) → +¹(0(*(x, y)), y)
+¹(j(x), j(y)) → +¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
*¹(1(x), y) → 0¹(*(x, y))
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
*¹(+(x, y), z) → *¹(y, z)
-¹(x, y) → +¹(x, opp(y))
+¹(j(x), 1(y)) → 0¹(+(x, y))
+¹(1(x), j(y)) → 0¹(+(x, y))
*¹(0(x), y) → *¹(x, y)
OPP(0(x)) → 0¹(opp(x))
OPP(0(x)) → OPP(x)
+¹(+(x, y), z) → +¹(y, z)
OPP(1(x)) → OPP(x)
*¹(j(x), y) → -¹(0(*(x, y)), y)
*¹(x, +(y, z)) → +¹(*(x, y), *(x, z))
+¹(0(x), 0(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(0(x), 1(y)) → +¹(x, y)
-¹(x, y) → OPP(y)
*¹(*(x, y), z) → *¹(y, z)
*¹(+(x, y), z) → +¹(*(x, z), *(y, z))
+¹(+(x, y), z) → +¹(x, +(y, z))
*¹(*(x, y), z) → *¹(x, *(y, z))
+¹(j(x), j(y)) → +¹(+(x, y), j(#))
*¹(j(x), y) → 0¹(*(x, y))
*¹(1(x), y) → *¹(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 13 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

          ↳ QDP

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

OPP(0(x)) → OPP(x)
OPP(j(x)) → OPP(x)
OPP(1(x)) → OPP(x)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ QDPSizeChangeProof

          ↳ QDP

          ↳ QDP

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

OPP(0(x)) → OPP(x)
OPP(j(x)) → OPP(x)
OPP(1(x)) → OPP(x)

R is empty.
Q is empty.
We have to consider all minimal (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:

OPP(0(x)) → OPP(x)
The graph contains the following edges 1 > 1
OPP(j(x)) → OPP(x)
The graph contains the following edges 1 > 1
OPP(1(x)) → OPP(x)
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

          ↳ QDP

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

+¹(0(x), j(y)) → +¹(x, y)
+¹(j(x), 0(y)) → +¹(x, y)
+¹(1(x), j(y)) → +¹(x, y)
+¹(j(x), 1(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(y, z)
+¹(1(x), 1(y)) → +¹(x, y)
+¹(0(x), 0(y)) → +¹(x, y)
+¹(0(x), 1(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(j(x), j(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
+¹(j(x), j(y)) → +¹(+(x, y), j(#))

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

          ↳ QDP

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

+¹(0(x), j(y)) → +¹(x, y)
+¹(j(x), 0(y)) → +¹(x, y)
+¹(1(x), j(y)) → +¹(x, y)
+¹(j(x), 1(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(y, z)
+¹(1(x), 1(y)) → +¹(x, y)
+¹(0(x), 0(y)) → +¹(x, y)
+¹(0(x), 1(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(j(x), j(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
+¹(j(x), j(y)) → +¹(+(x, y), j(#))

The TRS R consists of the following rules:

+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
0(#) → #

Q is empty.
We have to consider all minimal (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:

+¹(0(x), j(y)) → +¹(x, y)
+¹(j(x), 0(y)) → +¹(x, y)
+¹(1(x), j(y)) → +¹(x, y)
+¹(j(x), 1(y)) → +¹(x, y)
+¹(1(x), 1(y)) → +¹(x, y)
+¹(0(x), 0(y)) → +¹(x, y)
+¹(0(x), 1(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(j(x), j(y)) → +¹(x, y)
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
+¹(j(x), j(y)) → +¹(+(x, y), j(#))

Strictly oriented rules of the TRS R:

+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
0(#) → #

Used ordering: POLO with Polynomial interpretation [25]:

POL(#) = 0
POL(+(x₁, x₂)) = x₁ + x₂
POL(+¹(x₁, x₂)) = 2·x₁ + 2·x₂
POL(0(x₁)) = 1 + x₁
POL(1(x₁)) = 1 + x₁
POL(j(x₁)) = 1 + x₁

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ RuleRemovalProof

                  ↳ QDP

                    ↳ QDPSizeChangeProof

          ↳ QDP

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

+¹(+(x, y), z) → +¹(y, z)
+¹(+(x, y), z) → +¹(x, +(y, z))

The TRS R consists of the following rules:

+(#, x) → x
+(x, #) → x
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(+(x, y), z) → +(x, +(y, z))

Q is empty.
We have to consider all minimal (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:

+¹(+(x, y), z) → +¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
+¹(+(x, y), z) → +¹(x, +(y, z))
The graph contains the following edges 1 > 1

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPSizeChangeProof

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

*¹(x, +(y, z)) → *¹(x, y)
*¹(+(x, y), z) → *¹(x, z)
*¹(j(x), y) → *¹(x, y)
*¹(x, +(y, z)) → *¹(x, z)
*¹(*(x, y), z) → *¹(y, z)
*¹(*(x, y), z) → *¹(x, *(y, z))
*¹(+(x, y), z) → *¹(y, z)
*¹(0(x), y) → *¹(x, y)
*¹(1(x), y) → *¹(x, y)

The TRS R consists of the following rules:

0(#) → #
+(#, x) → x
+(x, #) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(0(x), j(y)) → j(+(x, y))
+(j(x), 0(y)) → j(+(x, y))
+(1(x), 1(y)) → j(+(+(x, y), 1(#)))
+(j(x), j(y)) → 1(+(+(x, y), j(#)))
+(1(x), j(y)) → 0(+(x, y))
+(j(x), 1(y)) → 0(+(x, y))
+(+(x, y), z) → +(x, +(y, z))
opp(#) → #
opp(0(x)) → 0(opp(x))
opp(1(x)) → j(opp(x))
opp(j(x)) → 1(opp(x))
-(x, y) → +(x, opp(y))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(j(x), y) → -(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
*(+(x, y), z) → +(*(x, z), *(y, z))
*(x, +(y, z)) → +(*(x, y), *(x, z))

From the DPs we obtained the following set of size-change graphs:

*¹(1(x), y) → *¹(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(0(x), y) → *¹(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(j(x), y) → *¹(x, y)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(*(x, y), z) → *¹(x, *(y, z))
The graph contains the following edges 1 > 1
*¹(*(x, y), z) → *¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(+(x, y), z) → *¹(x, z)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(+(x, y), z) → *¹(y, z)
The graph contains the following edges 1 > 1, 2 >= 2
*¹(x, +(y, z)) → *¹(x, y)
The graph contains the following edges 1 >= 1, 2 > 2
*¹(x, +(y, z)) → *¹(x, z)
The graph contains the following edges 1 >= 1, 2 > 2