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:

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(x1)

Q is empty.


↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(x1)

Q is empty.

We have reversed the following QTRS:
The set of rules R is

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(x1)

The set Q is empty.
We have obtained the following QTRS:

d(a(x)) → b(d(x))
a(x) → b(b(b(x)))
b(d(b(x))) → c(a(x))
c(x) → d(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
    ↳ QTRS
  ↳ DependencyPairsProof
  ↳ QTRS Reverse

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

d(a(x)) → b(d(x))
a(x) → b(b(b(x)))
b(d(b(x))) → c(a(x))
c(x) → d(x)

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:

A(d(x1)) → B(x1)
A(x1) → B(b(x1))
B(d(b(x1))) → C(x1)
A(x1) → B(b(b(x1)))
B(d(b(x1))) → A(c(x1))
A(x1) → B(x1)

The TRS R consists of the following rules:

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(x1)

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

A(d(x1)) → B(x1)
A(x1) → B(b(x1))
B(d(b(x1))) → C(x1)
A(x1) → B(b(b(x1)))
B(d(b(x1))) → A(c(x1))
A(x1) → B(x1)

The TRS R consists of the following rules:

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(x1)

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

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
  ↳ QTRS Reverse

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

A(d(x1)) → B(x1)
A(x1) → B(b(x1))
A(x1) → B(b(b(x1)))
B(d(b(x1))) → A(c(x1))
A(x1) → B(x1)

The TRS R consists of the following rules:

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(x1)

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:

A(d(x1)) → B(x1)


Used ordering: POLO with Polynomial interpretation [25]:

POL(A(x1)) = 2·x1   
POL(B(x1)) = 2·x1   
POL(a(x1)) = x1   
POL(b(x1)) = x1   
POL(c(x1)) = 1 + x1   
POL(d(x1)) = 1 + x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ QDPOrderProof
  ↳ QTRS Reverse

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

A(x1) → B(b(x1))
A(x1) → B(b(b(x1)))
B(d(b(x1))) → A(c(x1))
A(x1) → B(x1)

The TRS R consists of the following rules:

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(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.


A(x1) → B(b(x1))
B(d(b(x1))) → A(c(x1))
A(x1) → B(x1)
The remaining pairs can at least be oriented weakly.

A(x1) → B(b(b(x1)))
Used ordering: Polynomial interpretation [25]:

POL(A(x1)) = 2 + x1   
POL(B(x1)) = x1   
POL(a(x1)) = 8 + x1   
POL(b(x1)) = 1 + x1   
POL(c(x1)) = 1 + 8·x1   
POL(d(x1)) = 8·x1   

The following usable rules [17] were oriented:

a(d(x1)) → d(b(x1))
b(d(b(x1))) → a(c(x1))
a(x1) → b(b(b(x1)))
c(x1) → d(x1)



↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ RuleRemovalProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ DependencyGraphProof
  ↳ QTRS Reverse

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

A(x1) → B(b(b(x1)))

The TRS R consists of the following rules:

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(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.
We have reversed the following QTRS:
The set of rules R is

a(d(x1)) → d(b(x1))
a(x1) → b(b(b(x1)))
b(d(b(x1))) → a(c(x1))
c(x1) → d(x1)

The set Q is empty.
We have obtained the following QTRS:

d(a(x)) → b(d(x))
a(x) → b(b(b(x)))
b(d(b(x))) → c(a(x))
c(x) → d(x)

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ DependencyPairsProof
  ↳ QTRS Reverse
    ↳ QTRS

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

d(a(x)) → b(d(x))
a(x) → b(b(b(x)))
b(d(b(x))) → c(a(x))
c(x) → d(x)

Q is empty.