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:

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1

Q is empty.


QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1

Q is empty.

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1

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

0(sq(x)) → s(p(s(p(0(s(s(s(s(p(p(p(p(s(p(s(p(x)))))))))))))))))
s(sq(x)) → s(s(s(s(s(s(p(p(p(p(p(p(sq(s(s(s(p(p(p(s(p(s(p(twice(s(s(p(p(s(p(s(p(s(x)))))))))))))))))))))))))))))))))
0(twice(x)) → s(p(s(p(s(p(s(s(p(p(s(s(s(p(p(p(0(s(s(s(p(s(s(p(p(p(p(x)))))))))))))))))))))))))))
s(twice(x)) → s(p(s(p(twice(s(s(s(p(p(s(s(s(p(p(x)))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(s(s(s(s(s(s(s(0(x))))))))))))
0(x) → x

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
QTRS
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof

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

0(sq(x)) → s(p(s(p(0(s(s(s(s(p(p(p(p(s(p(s(p(x)))))))))))))))))
s(sq(x)) → s(s(s(s(s(s(p(p(p(p(p(p(sq(s(s(s(p(p(p(s(p(s(p(twice(s(s(p(p(s(p(s(p(s(x)))))))))))))))))))))))))))))))))
0(twice(x)) → s(p(s(p(s(p(s(s(p(p(s(s(s(p(p(p(0(s(s(s(p(s(s(p(p(p(p(x)))))))))))))))))))))))))))
s(twice(x)) → s(p(s(p(twice(s(s(s(p(p(s(s(s(p(p(x)))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(s(s(s(s(s(s(s(0(x))))))))))))
0(x) → x

Q is empty.

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1

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

0(sq(x)) → s(p(s(p(0(s(s(s(s(p(p(p(p(s(p(s(p(x)))))))))))))))))
s(sq(x)) → s(s(s(s(s(s(p(p(p(p(p(p(sq(s(s(s(p(p(p(s(p(s(p(twice(s(s(p(p(s(p(s(p(s(x)))))))))))))))))))))))))))))))))
0(twice(x)) → s(p(s(p(s(p(s(s(p(p(s(s(s(p(p(p(0(s(s(s(p(s(s(p(p(p(p(x)))))))))))))))))))))))))))
s(twice(x)) → s(p(s(p(twice(s(s(s(p(p(s(s(s(p(p(x)))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(s(s(s(s(s(s(s(0(x))))))))))))
0(x) → x

The set Q is empty.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
QTRS
  ↳ RRRPoloQTRSProof

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

0(sq(x)) → s(p(s(p(0(s(s(s(s(p(p(p(p(s(p(s(p(x)))))))))))))))))
s(sq(x)) → s(s(s(s(s(s(p(p(p(p(p(p(sq(s(s(s(p(p(p(s(p(s(p(twice(s(s(p(p(s(p(s(p(s(x)))))))))))))))))))))))))))))))))
0(twice(x)) → s(p(s(p(s(p(s(s(p(p(s(s(s(p(p(p(0(s(s(s(p(s(s(p(p(p(p(x)))))))))))))))))))))))))))
s(twice(x)) → s(p(s(p(twice(s(s(s(p(p(s(s(s(p(p(x)))))))))))))))
s(p(p(x))) → p(x)
s(p(x)) → x
0(p(x)) → s(s(s(s(s(s(s(s(s(s(s(0(x))))))))))))
0(x) → x

Q is empty.

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1

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

sq(0(x1)) → p(s(p(s(p(p(p(p(s(s(s(s(0(p(s(p(s(x1)))))))))))))))))
Used ordering:
Polynomial interpretation [25]:

POL(0(x1)) = 2·x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   
POL(sq(x1)) = 2 + 2·x1   
POL(twice(x1)) = x1   




↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
QTRS
      ↳ RRRPoloQTRSProof

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

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1

Q is empty.

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

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))
0(x1) → x1

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

0(x1) → x1
Used ordering:
Polynomial interpretation [25]:

POL(0(x1)) = 2 + 2·x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   
POL(sq(x1)) = 2·x1   
POL(twice(x1)) = x1   




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

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

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(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:

TWICE(0(x1)) → P(p(s(s(p(s(p(s(p(s(x1))))))))))
TWICE(0(x1)) → P(s(p(s(p(s(x1))))))
SQ(s(x1)) → P(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(s(s(s(x1)))))))))
TWICE(0(x1)) → P(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))
SQ(s(x1)) → TWICE(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))
TWICE(0(x1)) → P(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))
SQ(s(x1)) → P(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(twice(p(s(p(s(x1)))))))))
TWICE(0(x1)) → P(s(s(p(s(p(s(p(s(x1)))))))))
SQ(s(x1)) → P(p(p(p(p(s(s(s(s(s(s(x1)))))))))))
TWICE(0(x1)) → P(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))
TWICE(s(x1)) → P(s(x1))
P(p(s(x1))) → P(x1)
TWICE(0(x1)) → P(s(x1))
SQ(s(x1)) → P(p(s(s(s(s(s(s(x1))))))))
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
SQ(s(x1)) → P(p(p(p(s(s(s(s(s(s(x1))))))))))
TWICE(s(x1)) → P(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
TWICE(0(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(s(s(s(s(s(x1)))))))
TWICE(0(x1)) → P(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))
TWICE(0(x1)) → P(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))
SQ(s(x1)) → P(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
SQ(s(x1)) → P(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))
TWICE(s(x1)) → P(p(s(s(s(twice(p(s(p(s(x1))))))))))
TWICE(s(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))
SQ(s(x1)) → P(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))
SQ(s(x1)) → P(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))
TWICE(0(x1)) → P(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))

The TRS R consists of the following rules:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

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

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

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

TWICE(0(x1)) → P(p(s(s(p(s(p(s(p(s(x1))))))))))
TWICE(0(x1)) → P(s(p(s(p(s(x1))))))
SQ(s(x1)) → P(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(s(s(s(x1)))))))))
TWICE(0(x1)) → P(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))
SQ(s(x1)) → TWICE(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))
TWICE(0(x1)) → P(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → P(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))
SQ(s(x1)) → P(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(twice(p(s(p(s(x1)))))))))
TWICE(0(x1)) → P(s(s(p(s(p(s(p(s(x1)))))))))
SQ(s(x1)) → P(p(p(p(p(s(s(s(s(s(s(x1)))))))))))
TWICE(0(x1)) → P(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))
SQ(s(x1)) → P(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))
TWICE(s(x1)) → P(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1))))))))))))))
TWICE(s(x1)) → P(s(x1))
P(p(s(x1))) → P(x1)
TWICE(0(x1)) → P(s(x1))
SQ(s(x1)) → P(p(s(s(s(s(s(s(x1))))))))
TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))
SQ(s(x1)) → P(p(p(p(s(s(s(s(s(s(x1))))))))))
TWICE(s(x1)) → P(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
TWICE(0(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(s(s(s(s(s(x1)))))))
TWICE(0(x1)) → P(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))
TWICE(0(x1)) → P(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))
SQ(s(x1)) → P(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))
SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))
SQ(s(x1)) → P(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))))))))))))))
TWICE(0(x1)) → P(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))
TWICE(s(x1)) → P(p(s(s(s(twice(p(s(p(s(x1))))))))))
TWICE(s(x1)) → P(s(p(s(x1))))
SQ(s(x1)) → P(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))
SQ(s(x1)) → P(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))))))))
SQ(s(x1)) → P(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))
TWICE(0(x1)) → P(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1))))))))))))))))))))))))))

The TRS R consists of the following rules:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(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 3 SCCs with 35 less nodes.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                  ↳ QDP
                  ↳ 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:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
QDP
                    ↳ UsableRulesProof
                  ↳ QDP
                  ↳ QDP

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

P(p(s(x1))) → P(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
QDP
                        ↳ UsableRulesReductionPairsProof
                  ↳ QDP
                  ↳ QDP

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

P(p(s(x1))) → P(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

P(p(s(x1))) → P(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(P(x1)) = 2·x1   
POL(p(x1)) = 2·x1   
POL(s(x1)) = 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ UsableRulesProof
                    ↳ UsableRulesProof
                      ↳ QDP
                        ↳ UsableRulesReductionPairsProof
QDP
                            ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
QDP
                        ↳ UsableRulesProof
                    ↳ MNOCProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                            ↳ QReductionProof
                    ↳ MNOCProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

sq(s(x0))
twice(0(x0))
twice(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
QDP
                                ↳ QReductionProof
                    ↳ MNOCProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as they contain symbols which do neither occur in P nor in R.

p(0(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ QReductionProof
QDP
                                    ↳ MNOCProof
                    ↳ MNOCProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))

We have to consider all (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ QReductionProof
                                  ↳ QDP
                                    ↳ MNOCProof
QDP
                    ↳ MNOCProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

Q is empty.
We have to consider all (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
QDP
                        ↳ UsableRulesProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                            ↳ QReductionProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

sq(s(x0))
twice(0(x0))
twice(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
QDP
                                ↳ Rewriting
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(p(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule TWICE(s(x1)) → TWICE(p(s(p(s(x1))))) at position [0] we obtained the following new rules:

TWICE(s(x1)) → TWICE(p(s(x1)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
QDP
                                    ↳ Rewriting
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(p(s(x1)))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule TWICE(s(x1)) → TWICE(p(s(x1))) at position [0] we obtained the following new rules:

TWICE(s(x1)) → TWICE(x1)



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
QDP
                                        ↳ UsableRulesProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(x1)

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ UsableRulesProof
QDP
                                            ↳ QReductionProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(x1)

R is empty.
The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

p(s(x0))
p(0(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ QReductionProof
QDP
                                                ↳ UsableRulesReductionPairsProof
                  ↳ QDP

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

TWICE(s(x1)) → TWICE(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

TWICE(s(x1)) → TWICE(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(TWICE(x1)) = 2·x1   
POL(s(x1)) = 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ UsableRulesProof
                                          ↳ QDP
                                            ↳ QReductionProof
                                              ↳ QDP
                                                ↳ UsableRulesReductionPairsProof
QDP
                                                    ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
R is empty.
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
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
QDP
                    ↳ MNOCProof
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
QDP
                        ↳ UsableRulesProof
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                            ↳ QReductionProof
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

sq(s(x0))
twice(0(x0))
twice(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
QDP
                                ↳ Rewriting
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1))))))))))))) at position [0,0,0,0,0,0] we obtained the following new rules:

SQ(s(x1)) → SQ(p(p(p(p(p(s(s(s(s(s(x1)))))))))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
QDP
                                    ↳ Rewriting
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(p(p(p(s(s(s(s(s(x1)))))))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(p(p(p(s(s(s(s(s(x1))))))))))) at position [0,0,0,0,0] we obtained the following new rules:

SQ(s(x1)) → SQ(p(p(p(p(s(s(s(s(x1)))))))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
QDP
                                        ↳ Rewriting
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(p(p(s(s(s(s(x1)))))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(p(p(s(s(s(s(x1))))))))) at position [0,0,0,0] we obtained the following new rules:

SQ(s(x1)) → SQ(p(p(p(s(s(s(x1)))))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
QDP
                                            ↳ Rewriting
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(p(s(s(s(x1)))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(p(s(s(s(x1))))))) at position [0,0,0] we obtained the following new rules:

SQ(s(x1)) → SQ(p(p(s(s(x1)))))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
QDP
                                                ↳ Rewriting
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(s(s(x1)))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(p(s(s(x1))))) at position [0,0] we obtained the following new rules:

SQ(s(x1)) → SQ(p(s(x1)))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
QDP
                                                    ↳ UsableRulesProof
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(s(x1)))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
QDP
                                                        ↳ Rewriting
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(s(x1)))

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule SQ(s(x1)) → SQ(p(s(x1))) at position [0] we obtained the following new rules:

SQ(s(x1)) → SQ(x1)



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
                                                      ↳ QDP
                                                        ↳ Rewriting
QDP
                                                            ↳ UsableRulesProof
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(x1)

The TRS R consists of the following rules:

p(s(x1)) → x1

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ UsableRulesProof
QDP
                                                                ↳ QReductionProof
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(x1)

R is empty.
The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

p(s(x0))
p(0(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ UsableRulesProof
                                                              ↳ QDP
                                                                ↳ QReductionProof
QDP
                                                                    ↳ UsableRulesReductionPairsProof
                    ↳ MNOCProof

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

SQ(s(x1)) → SQ(x1)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

SQ(s(x1)) → SQ(x1)
No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [25]:

POL(SQ(x1)) = 2·x1   
POL(s(x1)) = 2·x1   



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ Rewriting
                                  ↳ QDP
                                    ↳ Rewriting
                                      ↳ QDP
                                        ↳ Rewriting
                                          ↳ QDP
                                            ↳ Rewriting
                                              ↳ QDP
                                                ↳ Rewriting
                                                  ↳ QDP
                                                    ↳ UsableRulesProof
                                                      ↳ QDP
                                                        ↳ Rewriting
                                                          ↳ QDP
                                                            ↳ UsableRulesProof
                                                              ↳ QDP
                                                                ↳ QReductionProof
                                                                  ↳ QDP
                                                                    ↳ UsableRulesReductionPairsProof
QDP
                                                                        ↳ PisEmptyProof
                    ↳ MNOCProof

Q DP problem:
P is empty.
R is empty.
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.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
QDP
                        ↳ UsableRulesProof

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

sq(s(x1)) → s(p(s(p(s(p(p(s(s(twice(p(s(p(s(p(p(p(s(s(s(sq(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))))))))))))))))))))))
twice(0(x1)) → p(p(p(p(s(s(p(s(s(s(0(p(p(p(s(s(s(p(p(s(s(p(s(p(s(p(s(x1)))))))))))))))))))))))))))
twice(s(x1)) → p(p(s(s(s(p(p(s(s(s(twice(p(s(p(s(x1)))))))))))))))
p(p(s(x1))) → p(x1)
p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
QDP
                            ↳ QReductionProof

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

sq(s(x0))
twice(0(x0))
twice(s(x0))
p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

sq(s(x0))
twice(0(x0))
twice(s(x0))



↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
QDP
                                ↳ MNOCProof

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

The set Q consists of the following terms:

p(s(x0))
p(0(x0))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ QTRS Reverse
  ↳ QTRS Reverse
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ RRRPoloQTRSProof
        ↳ QTRS
          ↳ DependencyPairsProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP
                    ↳ MNOCProof
                    ↳ MNOCProof
                      ↳ QDP
                        ↳ UsableRulesProof
                          ↳ QDP
                            ↳ QReductionProof
                              ↳ QDP
                                ↳ MNOCProof
QDP

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

SQ(s(x1)) → SQ(p(p(p(p(p(p(s(s(s(s(s(s(x1)))))))))))))

The TRS R consists of the following rules:

p(s(x1)) → x1
p(0(x1)) → 0(s(s(s(s(s(s(s(s(s(s(s(x1))))))))))))

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