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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

01(0(*(*(x1)))) → 11(1(x1))
11(1(*(*(x1)))) → 01(0(#(#(x1))))
Used ordering: Polynomial interpretation [25,35]:

POL(*(x1)) = x_1   
POL(11(x1)) = 3/2 + (1/2)x_1   
POL(1(x1)) = 1/4 + (4)x_1   
POL(01(x1)) = 3/2 + (1/2)x_1   
POL(#1(x1)) = 5/4 + (2)x_1   
POL(0(x1)) = 1/4 + (4)x_1   
POL(#(x1)) = x_1   
POL($(x1)) = (4)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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



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

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

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

The TRS R consists of the following rules:

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


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

11(1(*(*(x1)))) → 01(0(#(#(x1))))
Used ordering: Polynomial interpretation [25,35]:

POL(11(x1)) = (1/4)x_1   
POL(1(x1)) = x_1   
POL(*(x1)) = 1/4 + x_1   
POL(01(x1)) = (1/4)x_1   
POL(0(x1)) = x_1   
POL(#(x1)) = 1/4 + x_1   
POL($(x1)) = 0   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

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



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

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

11(1(*(*(x1)))) → 01(0(#(#(x1))))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.