Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(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:

FOO(s(x1)) → P(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))
FOO(0(x1)) → P(p(p(s(s(s(p(s(x1))))))))
FOO(s(x1)) → P(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))
FOO(s(x1)) → P(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))))))))
FOO(s(x1)) → P(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))
BAR(s(x1)) → P(p(s(s(x1))))
FOO(s(x1)) → P(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))))))))))))
FOO(s(x1)) → P(s(s(p(s(x1)))))
FOO(s(x1)) → P(s(bar(p(p(s(s(p(s(x1)))))))))
P(p(s(x1))) → P(x1)
BAR(s(x1)) → P(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
FOO(0(x1)) → P(p(s(s(s(p(s(x1)))))))
FOO(0(x1)) → P(s(x1))
FOO(s(x1)) → P(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
FOO(s(x1)) → P(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))
FOO(s(x1)) → P(p(s(s(p(s(x1))))))
BAR(0(x1)) → P(s(s(s(x1))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))
FOO(s(x1)) → P(s(x1))
FOO(s(x1)) → P(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))
FOO(0(x1)) → P(s(s(s(p(s(x1))))))
BAR(s(x1)) → P(p(s(s(foo(s(p(p(s(s(x1))))))))))
FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
BAR(s(x1)) → P(s(s(foo(s(p(p(s(s(x1)))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → P(s(s(x1)))

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

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:

FOO(s(x1)) → P(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))
FOO(0(x1)) → P(p(p(s(s(s(p(s(x1))))))))
FOO(s(x1)) → P(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))
FOO(s(x1)) → P(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))))))))
FOO(s(x1)) → P(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))
BAR(s(x1)) → P(p(s(s(x1))))
FOO(s(x1)) → P(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1)))))))))))))))))))))))
FOO(s(x1)) → P(s(s(p(s(x1)))))
FOO(s(x1)) → P(s(bar(p(p(s(s(p(s(x1)))))))))
P(p(s(x1))) → P(x1)
BAR(s(x1)) → P(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
FOO(0(x1)) → P(p(s(s(s(p(s(x1)))))))
FOO(0(x1)) → P(s(x1))
FOO(s(x1)) → P(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
FOO(s(x1)) → P(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))
FOO(s(x1)) → P(p(s(s(p(s(x1))))))
BAR(0(x1)) → P(s(s(s(x1))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))
FOO(s(x1)) → P(s(x1))
FOO(s(x1)) → P(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))
FOO(0(x1)) → P(s(s(s(p(s(x1))))))
BAR(s(x1)) → P(p(s(s(foo(s(p(p(s(s(x1))))))))))
FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
BAR(s(x1)) → P(s(s(foo(s(p(p(s(s(x1)))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → P(s(s(x1)))

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 22 less nodes.

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

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

P(p(s(x1))) → P(x1)

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(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.


P(p(s(x1))) → P(x1)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = (2)x_1   
POL(p(x1)) = 2 + (2)x_1   
POL(s(x1)) = x_1   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

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

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP

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

FOO(s(x1)) → FOO(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))
FOO(s(x1)) → BAR(p(p(s(s(p(s(x1)))))))
BAR(s(x1)) → FOO(s(p(p(s(s(x1))))))

The TRS R consists of the following rules:

foo(0(x1)) → 0(s(p(p(p(s(s(s(p(s(x1))))))))))
foo(s(x1)) → p(s(p(p(p(s(s(p(s(s(p(s(foo(p(p(s(s(p(s(bar(p(p(s(s(p(s(x1))))))))))))))))))))))))))
bar(0(x1)) → 0(p(s(s(s(x1)))))
bar(s(x1)) → p(s(p(p(s(s(foo(s(p(p(s(s(x1))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(x1)))))

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