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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

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)
DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
ACTIVATE1(n__0) -> 01
ACTIVATE1(n__s1(X)) -> S1(X)
LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)
IF3(false, X, Y) -> ACTIVATE1(Y)
DIFF2(X, Y) -> P1(X)
IF3(true, X, Y) -> ACTIVATE1(X)
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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

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)
DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
ACTIVATE1(n__0) -> 01
ACTIVATE1(n__s1(X)) -> S1(X)
LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)
IF3(false, X, Y) -> ACTIVATE1(Y)
DIFF2(X, Y) -> P1(X)
IF3(true, X, Y) -> ACTIVATE1(X)
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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

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 7 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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

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 Order [17,21] with Interpretation:

POL( LEQ2(x1, x2) ) = x2


POL( s1(x1) ) = x1 + 1



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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(diff2(p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(X)
activate1(X) -> X

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