YES Termination w.r.t. Q proof of ../tpdb/TRS/Cime/dpqs.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:

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(0)) → g(d(1))
g(c(1)) → g(d(0))

Q is empty.


↳ QTRS
  ↳ RRRPoloQTRSProof

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

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(0)) → g(d(1))
g(c(1)) → g(d(0))

Q is empty.

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

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(x)) → x
g(d(x)) → x
g(c(0)) → g(d(1))
g(c(1)) → g(d(0))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

g(c(x)) → x
g(d(x)) → x
Used ordering:
Polynomial interpretation [25]:

POL(0) = 0   
POL(1) = 0   
POL(c(x1)) = x1   
POL(d(x1)) = x1   
POL(f(x1)) = 2·x1   
POL(g(x1)) = 1 + 2·x1   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof

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

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(0)) → g(d(1))
g(c(1)) → g(d(0))

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:

F(f(x)) → F(c(f(x)))
G(c(1)) → G(d(0))
F(f(x)) → F(d(f(x)))
G(c(0)) → G(d(1))

The TRS R consists of the following rules:

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(0)) → g(d(1))
g(c(1)) → g(d(0))

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

↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

F(f(x)) → F(c(f(x)))
G(c(1)) → G(d(0))
F(f(x)) → F(d(f(x)))
G(c(0)) → G(d(1))

The TRS R consists of the following rules:

f(f(x)) → f(c(f(x)))
f(f(x)) → f(d(f(x)))
g(c(0)) → g(d(1))
g(c(1)) → g(d(0))

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 4 less nodes.