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:

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))

Q is empty.

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

P1(p1(s1(x))) -> P1(x)
MINUS2(x, y) -> LE2(x, y)
MINUS2(x, y) -> IF3(le2(x, y), x, y)
LE2(p1(s1(x)), x) -> LE2(x, x)
LE2(s1(x), s1(y)) -> LE2(x, y)
IF3(false, x, y) -> MINUS2(p1(x), y)
IF3(false, x, y) -> P1(x)

The TRS R consists of the following rules:

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), 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:

P1(p1(s1(x))) -> P1(x)
MINUS2(x, y) -> LE2(x, y)
MINUS2(x, y) -> IF3(le2(x, y), x, y)
LE2(p1(s1(x)), x) -> LE2(x, x)
LE2(s1(x), s1(y)) -> LE2(x, y)
IF3(false, x, y) -> MINUS2(p1(x), y)
IF3(false, x, y) -> P1(x)

The TRS R consists of the following rules:

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 2 less nodes.

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

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

LE2(p1(s1(x)), x) -> LE2(x, x)
LE2(s1(x), s1(y)) -> LE2(x, y)

The TRS R consists of the following rules:

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE2(s1(x), s1(y)) -> LE2(x, y)
Used argument filtering: LE2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


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

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

LE2(p1(s1(x)), x) -> LE2(x, x)

The TRS R consists of the following rules:

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

LE2(p1(s1(x)), x) -> LE2(x, x)
Used argument filtering: LE2(x1, x2)  =  x1
p1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

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

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), 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.

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

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

P1(p1(s1(x))) -> P1(x)

The TRS R consists of the following rules:

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

P1(p1(s1(x))) -> P1(x)
Used argument filtering: P1(x1)  =  x1
p1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ PisEmptyProof
          ↳ QDP

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

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), 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.

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

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

MINUS2(x, y) -> IF3(le2(x, y), x, y)
IF3(false, x, y) -> MINUS2(p1(x), y)

The TRS R consists of the following rules:

p1(0) -> s1(s1(0))
p1(s1(x)) -> x
p1(p1(s1(x))) -> p1(x)
le2(p1(s1(x)), x) -> le2(x, x)
le2(0, y) -> true
le2(s1(x), 0) -> false
le2(s1(x), s1(y)) -> le2(x, y)
minus2(x, y) -> if3(le2(x, y), x, y)
if3(true, x, y) -> 0
if3(false, x, y) -> s1(minus2(p1(x), y))

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