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(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y

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:

C(b(y, c(x))) → A(0, 0)
C(b(y, c(x))) → C(c(b(a(0, 0), y)))
A(x, y) → C(y)
C(b(y, c(x))) → B(a(0, 0), y)
C(b(y, c(x))) → C(b(a(0, 0), y))
A(x, y) → B(0, c(y))
A(x, y) → B(x, b(0, c(y)))

The TRS R consists of the following rules:

a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y

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:

C(b(y, c(x))) → A(0, 0)
C(b(y, c(x))) → C(c(b(a(0, 0), y)))
A(x, y) → C(y)
C(b(y, c(x))) → B(a(0, 0), y)
C(b(y, c(x))) → C(b(a(0, 0), y))
A(x, y) → B(0, c(y))
A(x, y) → B(x, b(0, c(y)))

The TRS R consists of the following rules:

a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y

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 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Instantiation

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

C(b(y, c(x))) → A(0, 0)
C(b(y, c(x))) → C(c(b(a(0, 0), y)))
A(x, y) → C(y)
C(b(y, c(x))) → C(b(a(0, 0), y))

The TRS R consists of the following rules:

a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule A(x, y) → C(y) we obtained the following new rules:

A(0, 0) → C(0)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Instantiation
            ↳ QDP
              ↳ DependencyGraphProof

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

C(b(y, c(x))) → A(0, 0)
A(0, 0) → C(0)
C(b(y, c(x))) → C(c(b(a(0, 0), y)))
C(b(y, c(x))) → C(b(a(0, 0), y))

The TRS R consists of the following rules:

a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y

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 2 less nodes.

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

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

C(b(y, c(x))) → C(c(b(a(0, 0), y)))
C(b(y, c(x))) → C(b(a(0, 0), y))

The TRS R consists of the following rules:

a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y

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.


C(b(y, c(x))) → C(c(b(a(0, 0), y)))
C(b(y, c(x))) → C(b(a(0, 0), y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( c(x1) ) =
/1\
\0/
+
/01\
\00/
·x1

M( b(x1, x2) ) =
/0\
\0/
+
/10\
\11/
·x1+
/00\
\10/
·x2

M( a(x1, x2) ) =
/0\
\0/
+
/10\
\11/
·x1+
/00\
\00/
·x2

M( 0 ) =
/0\
\0/

Tuple symbols:
M( C(x1) ) = 0+
[0,1]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

c(b(y, c(x))) → c(c(b(a(0, 0), y)))
a(x, y) → b(x, b(0, c(y)))
b(y, 0) → y



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ Instantiation
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

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

a(x, y) → b(x, b(0, c(y)))
c(b(y, c(x))) → c(c(b(a(0, 0), y)))
b(y, 0) → y

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.