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(*(x1)) → *(1(x1))
1(*(x1)) → 0(#(x1))
#(0(x1)) → 0(#(x1))
#(1(x1)) → 1(#(x1))
#($(x1)) → *($(x1))
#(#(x1)) → #(x1)
#(*(x1)) → *(x1)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

0(*(x1)) → *(1(x1))
1(*(x1)) → 0(#(x1))
#(0(x1)) → 0(#(x1))
#(1(x1)) → 1(#(x1))
#($(x1)) → *($(x1))
#(#(x1)) → #(x1)
#(*(x1)) → *(x1)

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(0(x1)) → 01(#(x1))
11(*(x1)) → 01(#(x1))
11(*(x1)) → #1(x1)
#1(1(x1)) → 11(#(x1))
01(*(x1)) → 11(x1)
#1(0(x1)) → #1(x1)
#1(1(x1)) → #1(x1)

The TRS R consists of the following rules:

0(*(x1)) → *(1(x1))
1(*(x1)) → 0(#(x1))
#(0(x1)) → 0(#(x1))
#(1(x1)) → 1(#(x1))
#($(x1)) → *($(x1))
#(#(x1)) → #(x1)
#(*(x1)) → *(x1)

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

#1(0(x1)) → 01(#(x1))
11(*(x1)) → 01(#(x1))
11(*(x1)) → #1(x1)
#1(1(x1)) → 11(#(x1))
01(*(x1)) → 11(x1)
#1(0(x1)) → #1(x1)
#1(1(x1)) → #1(x1)

The TRS R consists of the following rules:

0(*(x1)) → *(1(x1))
1(*(x1)) → 0(#(x1))
#(0(x1)) → 0(#(x1))
#(1(x1)) → 1(#(x1))
#($(x1)) → *($(x1))
#(#(x1)) → #(x1)
#(*(x1)) → *(x1)

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


The following pairs can be oriented strictly and are deleted.


#1(0(x1)) → 01(#(x1))
11(*(x1)) → #1(x1)
#1(1(x1)) → 11(#(x1))
#1(0(x1)) → #1(x1)
#1(1(x1)) → #1(x1)
The remaining pairs can at least be oriented weakly.

11(*(x1)) → 01(#(x1))
01(*(x1)) → 11(x1)
Used ordering: Polynomial interpretation [25,35]:

POL(11(x1)) = 1 + (2)x_1   
POL(*(x1)) = x_1   
POL(1(x1)) = 2 + (4)x_1   
POL(01(x1)) = 1 + (2)x_1   
POL(#1(x1)) = x_1   
POL(0(x1)) = 2 + (4)x_1   
POL(#(x1)) = x_1   
POL($(x1)) = 0   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

0(*(x1)) → *(1(x1))
1(*(x1)) → 0(#(x1))
#(0(x1)) → 0(#(x1))
#(1(x1)) → 1(#(x1))
#($(x1)) → *($(x1))
#(#(x1)) → #(x1)
#(*(x1)) → *(x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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

11(*(x1)) → 01(#(x1))
01(*(x1)) → 11(x1)

The TRS R consists of the following rules:

0(*(x1)) → *(1(x1))
1(*(x1)) → 0(#(x1))
#(0(x1)) → 0(#(x1))
#(1(x1)) → 1(#(x1))
#($(x1)) → *($(x1))
#(#(x1)) → #(x1)
#(*(x1)) → *(x1)

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


The following pairs can be oriented strictly and are deleted.


11(*(x1)) → 01(#(x1))
01(*(x1)) → 11(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(11(x1)) = 1 + (4)x_1   
POL(*(x1)) = 1 + (4)x_1   
POL(1(x1)) = x_1   
POL(01(x1)) = 1 + x_1   
POL(0(x1)) = x_1   
POL(#(x1)) = 1 + (2)x_1   
POL($(x1)) = 0   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

0(*(x1)) → *(1(x1))
1(*(x1)) → 0(#(x1))
#(0(x1)) → 0(#(x1))
#(1(x1)) → 1(#(x1))
#($(x1)) → *($(x1))
#(#(x1)) → #(x1)
#(*(x1)) → *(x1)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

0(*(x1)) → *(1(x1))
1(*(x1)) → 0(#(x1))
#(0(x1)) → 0(#(x1))
#(1(x1)) → 1(#(x1))
#($(x1)) → *($(x1))
#(#(x1)) → #(x1)
#(*(x1)) → *(x1)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.