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
leq2(0, Y) -> true
leq2(s1(X), 0) -> false
leq2(s1(X), s1(Y)) -> leq2(X, Y)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
diff2(X, Y) -> if3(leq2(X, Y), 0, s1(diff2(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
leq2(0, Y) -> true
leq2(s1(X), 0) -> false
leq2(s1(X), s1(Y)) -> leq2(X, Y)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
diff2(X, Y) -> if3(leq2(X, Y), 0, s1(diff2(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:

DIFF2(X, Y) -> LEQ2(X, Y)
LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)
DIFF2(X, Y) -> P1(X)
DIFF2(X, Y) -> DIFF2(p1(X), Y)
DIFF2(X, Y) -> IF3(leq2(X, Y), 0, s1(diff2(p1(X), Y)))

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(X)) -> X
leq2(0, Y) -> true
leq2(s1(X), 0) -> false
leq2(s1(X), s1(Y)) -> leq2(X, Y)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
diff2(X, Y) -> if3(leq2(X, Y), 0, s1(diff2(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:

DIFF2(X, Y) -> LEQ2(X, Y)
LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)
DIFF2(X, Y) -> P1(X)
DIFF2(X, Y) -> DIFF2(p1(X), Y)
DIFF2(X, Y) -> IF3(leq2(X, Y), 0, s1(diff2(p1(X), Y)))

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(X)) -> X
leq2(0, Y) -> true
leq2(s1(X), 0) -> false
leq2(s1(X), s1(Y)) -> leq2(X, Y)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
diff2(X, Y) -> if3(leq2(X, Y), 0, s1(diff2(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 3 less nodes.

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

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

LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(X)) -> X
leq2(0, Y) -> true
leq2(s1(X), 0) -> false
leq2(s1(X), s1(Y)) -> leq2(X, Y)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
diff2(X, Y) -> if3(leq2(X, Y), 0, s1(diff2(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.


LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(LEQ2(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
leq2(0, Y) -> true
leq2(s1(X), 0) -> false
leq2(s1(X), s1(Y)) -> leq2(X, Y)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
diff2(X, Y) -> if3(leq2(X, Y), 0, s1(diff2(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:

DIFF2(X, Y) -> DIFF2(p1(X), Y)

The TRS R consists of the following rules:

p1(0) -> 0
p1(s1(X)) -> X
leq2(0, Y) -> true
leq2(s1(X), 0) -> false
leq2(s1(X), s1(Y)) -> leq2(X, Y)
if3(true, X, Y) -> X
if3(false, X, Y) -> Y
diff2(X, Y) -> if3(leq2(X, Y), 0, s1(diff2(p1(X), Y)))

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