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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
leq2(0, x0)
leq2(s1(x0), 0)
leq2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
diff2(x0, x1)


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)))

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
leq2(0, x0)
leq2(s1(x0), 0)
leq2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
diff2(x0, x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
leq2(0, x0)
leq2(s1(x0), 0)
leq2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
diff2(x0, x1)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ 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)))

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
leq2(0, x0)
leq2(s1(x0), 0)
leq2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
diff2(x0, x1)

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.

LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)
Used argument filtering: LEQ2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ 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)))

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
leq2(0, x0)
leq2(s1(x0), 0)
leq2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
diff2(x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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)))

The set Q consists of the following terms:

p1(0)
p1(s1(x0))
leq2(0, x0)
leq2(s1(x0), 0)
leq2(s1(x0), s1(x1))
if3(true, x0, x1)
if3(false, x0, x1)
diff2(x0, x1)

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