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) -> 0
p1(s1(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) -> 0
p1(s1(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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MINUS2(x, y) -> LE2(x, y)
MINUS2(x, y) -> IF3(le2(x, y), x, y)
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) -> 0
p1(s1(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:

MINUS2(x, y) -> LE2(x, y)
MINUS2(x, y) -> IF3(le2(x, y), x, y)
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) -> 0
p1(s1(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 [13,14,18] contains 2 SCCs with 2 less nodes.

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

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

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

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


LE2(s1(x), s1(y)) -> LE2(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LE2(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] 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:

p1(0) -> 0
p1(s1(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

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) -> 0
p1(s1(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.