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(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(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(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(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:

ACTIVATE1(n__0) -> 01
DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> DIFF2(activate1(X1), activate1(X2))
DIFF2(X, Y) -> LEQ2(X, Y)
ACTIVATE1(n__diff2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> ACTIVATE1(X2)
LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)
IF3(false, X, Y) -> ACTIVATE1(Y)
ACTIVATE1(n__p1(X)) -> P1(activate1(X))
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

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(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(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:

ACTIVATE1(n__0) -> 01
DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> DIFF2(activate1(X1), activate1(X2))
DIFF2(X, Y) -> LEQ2(X, Y)
ACTIVATE1(n__diff2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__s1(X)) -> S1(activate1(X))
ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> ACTIVATE1(X2)
LEQ2(s1(X), s1(Y)) -> LEQ2(X, Y)
IF3(false, X, Y) -> ACTIVATE1(Y)
ACTIVATE1(n__p1(X)) -> P1(activate1(X))
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

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(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(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 4 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(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(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 interpretation [21]:

POL(LEQ2(x1, x2)) = 3·x1 + 3·x2   
POL(s1(x1)) = 2 + 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) -> activate1(X)
if3(false, X, Y) -> activate1(Y)
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(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
            ↳ QDPOrderProof

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

ACTIVATE1(n__diff2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> ACTIVATE1(X2)
DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
IF3(false, X, Y) -> ACTIVATE1(Y)
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> DIFF2(activate1(X1), activate1(X2))
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

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(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(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.


ACTIVATE1(n__diff2(X1, X2)) -> ACTIVATE1(X1)
ACTIVATE1(n__diff2(X1, X2)) -> ACTIVATE1(X2)
The remaining pairs can at least be oriented weakly.

ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
IF3(false, X, Y) -> ACTIVATE1(Y)
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> DIFF2(activate1(X1), activate1(X2))
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(ACTIVATE1(x1)) = 1 + x1   
POL(DIFF2(x1, x2)) = 2 + x1 + 3·x2   
POL(IF3(x1, x2, x3)) = 1 + 2·x2 + x3   
POL(activate1(x1)) = x1   
POL(diff2(x1, x2)) = 1 + x1 + 3·x2   
POL(false) = 0   
POL(if3(x1, x2, x3)) = x2 + x3   
POL(leq2(x1, x2)) = 0   
POL(n__0) = 0   
POL(n__diff2(x1, x2)) = 1 + x1 + 3·x2   
POL(n__p1(x1)) = x1   
POL(n__s1(x1)) = x1   
POL(p1(x1)) = x1   
POL(s1(x1)) = x1   
POL(true) = 0   

The following usable rules [14] were oriented:

s1(X) -> n__s1(X)
p1(s1(X)) -> X
p1(0) -> 0
activate1(n__p1(X)) -> p1(activate1(X))
diff2(X1, X2) -> n__diff2(X1, X2)
if3(false, X, Y) -> activate1(Y)
if3(true, X, Y) -> activate1(X)
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
diff2(X, Y) -> if3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
p1(X) -> n__p1(X)
0 -> n__0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(X) -> X
activate1(n__0) -> 0



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

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

ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
IF3(false, X, Y) -> ACTIVATE1(Y)
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> DIFF2(activate1(X1), activate1(X2))

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(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(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.


ACTIVATE1(n__p1(X)) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.

DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
IF3(false, X, Y) -> ACTIVATE1(Y)
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> DIFF2(activate1(X1), activate1(X2))
Used ordering: Polynomial interpretation [21]:

POL(0) = 3   
POL(ACTIVATE1(x1)) = 1 + x1   
POL(DIFF2(x1, x2)) = 1   
POL(IF3(x1, x2, x3)) = 1 + x2 + 2·x3   
POL(activate1(x1)) = 2·x1   
POL(diff2(x1, x2)) = 3·x1   
POL(false) = 0   
POL(if3(x1, x2, x3)) = 0   
POL(leq2(x1, x2)) = 2·x1   
POL(n__0) = 0   
POL(n__diff2(x1, x2)) = 0   
POL(n__p1(x1)) = 2 + x1   
POL(n__s1(x1)) = 2·x1   
POL(p1(x1)) = 0   
POL(s1(x1)) = 2·x1   
POL(true) = 0   

The following usable rules [14] were oriented: none



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

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

DIFF2(X, Y) -> IF3(leq2(X, Y), n__0, n__s1(n__diff2(n__p1(X), Y)))
IF3(false, X, Y) -> ACTIVATE1(Y)
IF3(true, X, Y) -> ACTIVATE1(X)
ACTIVATE1(n__diff2(X1, X2)) -> DIFF2(activate1(X1), activate1(X2))
ACTIVATE1(n__s1(X)) -> ACTIVATE1(X)

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(n__diff2(n__p1(X), Y)))
0 -> n__0
s1(X) -> n__s1(X)
diff2(X1, X2) -> n__diff2(X1, X2)
p1(X) -> n__p1(X)
activate1(n__0) -> 0
activate1(n__s1(X)) -> s1(activate1(X))
activate1(n__diff2(X1, X2)) -> diff2(activate1(X1), activate1(X2))
activate1(n__p1(X)) -> p1(activate1(X))
activate1(X) -> X

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