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:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MOD(s(x), s(y)) → LEQ(y, x)
-1(s(x), s(y)) → -1(x, y)
MOD(s(x), s(y)) → MOD(-(s(x), s(y)), s(y))
LEQ(s(x), s(y)) → LEQ(x, y)
MOD(s(x), s(y)) → IF(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))
MOD(s(x), s(y)) → -1(s(x), s(y))

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

MOD(s(x), s(y)) → LEQ(y, x)
-1(s(x), s(y)) → -1(x, y)
MOD(s(x), s(y)) → MOD(-(s(x), s(y)), s(y))
LEQ(s(x), s(y)) → LEQ(x, y)
MOD(s(x), s(y)) → IF(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))
MOD(s(x), s(y)) → -1(s(x), s(y))

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

MOD(s(x), s(y)) → LEQ(y, x)
-1(s(x), s(y)) → -1(x, y)
LEQ(s(x), s(y)) → LEQ(x, y)
MOD(s(x), s(y)) → MOD(-(s(x), s(y)), s(y))
MOD(s(x), s(y)) → IF(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))
MOD(s(x), s(y)) → -1(s(x), s(y))

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP
                  ↳ QDP

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

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))

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.


-1(s(x), s(y)) → -1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
-1(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP
                  ↳ QDP

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

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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

LEQ(s(x), s(y)) → LEQ(x, y)

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))

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.


LEQ(s(x), s(y)) → LEQ(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LEQ(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [2].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof
                  ↳ QDP

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

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                  ↳ QDP

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

MOD(s(x), s(y)) → MOD(-(s(x), s(y)), s(y))

The TRS R consists of the following rules:

leq(0, y) → true
leq(s(x), 0) → false
leq(s(x), s(y)) → leq(x, y)
if(true, x, y) → x
if(false, x, y) → y
-(x, 0) → x
-(s(x), s(y)) → -(x, y)
mod(0, y) → 0
mod(s(x), 0) → 0
mod(s(x), s(y)) → if(leq(y, x), mod(-(s(x), s(y)), s(y)), s(x))

The set Q consists of the following terms:

leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
-(x0, 0)
-(s(x0), s(x1))
mod(0, x0)
mod(s(x0), 0)
mod(s(x0), s(x1))

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