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:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(x))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(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:

MOD2(s1(x), s1(y)) -> LEQ2(y, x)
-12(s1(x), s1(y)) -> -12(x, y)
MOD2(s1(x), s1(y)) -> MOD2(-2(s1(x), s1(y)), s1(y))
LEQ2(s1(x), s1(y)) -> LEQ2(x, y)
MOD2(s1(x), s1(y)) -> IF3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(x))
MOD2(s1(x), s1(y)) -> -12(s1(x), s1(y))

The TRS R consists of the following rules:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(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:

MOD2(s1(x), s1(y)) -> LEQ2(y, x)
-12(s1(x), s1(y)) -> -12(x, y)
MOD2(s1(x), s1(y)) -> MOD2(-2(s1(x), s1(y)), s1(y))
LEQ2(s1(x), s1(y)) -> LEQ2(x, y)
MOD2(s1(x), s1(y)) -> IF3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(x))
MOD2(s1(x), s1(y)) -> -12(s1(x), s1(y))

The TRS R consists of the following rules:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

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

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

-12(s1(x), s1(y)) -> -12(x, y)

The TRS R consists of the following rules:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(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.


-12(s1(x), s1(y)) -> -12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s1(x1) ) = 2x1 + 2


POL( -12(x1, x2) ) = max{0, 2x1 + 2x2 - 1}



The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(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
          ↳ 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:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(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( s1(x1) ) = 2x1 + 2


POL( LEQ2(x1, x2) ) = max{0, 2x1 + 2x2 - 1}



The following usable rules [14] were oriented: none



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(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
          ↳ QDP

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

MOD2(s1(x), s1(y)) -> MOD2(-2(s1(x), s1(y)), s1(y))

The TRS R consists of the following rules:

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
-2(x, 0) -> x
-2(s1(x), s1(y)) -> -2(x, y)
mod2(0, y) -> 0
mod2(s1(x), 0) -> 0
mod2(s1(x), s1(y)) -> if3(leq2(y, x), mod2(-2(s1(x), s1(y)), s1(y)), s1(x))

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