Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)


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

F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
-1(s(x), s(y)) → -1(x, y)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
<=1(s(x), s(y)) → <=1(x, y)
F(s(x), 0, z, u) → -1(z, s(x))
F(s(x), s(y), z, u) → <=1(x, y)
F(s(x), s(y), z, u) → -1(y, x)
F(s(x), s(y), z, u) → IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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:

F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
-1(s(x), s(y)) → -1(x, y)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
<=1(s(x), s(y)) → <=1(x, y)
F(s(x), 0, z, u) → -1(z, s(x))
F(s(x), s(y), z, u) → <=1(x, y)
F(s(x), s(y), z, u) → -1(y, x)
F(s(x), s(y), z, u) → IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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:

F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
F(s(x), s(y), z, u) → F(x, u, z, u)
-1(s(x), s(y)) → -1(x, y)
<=1(s(x), s(y)) → <=1(x, y)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) → -1(z, s(x))
F(s(x), s(y), z, u) → <=1(x, y)
F(s(x), s(y), z, u) → IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) → -1(y, x)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 5 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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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)

Lexicographic Path Order [19].
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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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:

-1(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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)

Lexicographic Path Order [19].
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:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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
                    ↳ QDPOrderProof

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

F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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.


F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, -(z, s(x)), u)
The remaining pairs can at least be oriented weakly.

F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



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

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

F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)

The TRS R consists of the following rules:

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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.


F(s(x), s(y), z, u) → F(s(x), -(y, x), z, u)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4)  =  x2
s(x1)  =  s(x1)
-(x1, x2)  =  x1

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented:

-(s(x), s(y)) → -(x, y)
-(x, 0) → x



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

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

-(x, 0) → x
-(s(x), s(y)) → -(x, y)
<=(0, y) → true
<=(s(x), 0) → false
<=(s(x), s(y)) → <=(x, y)
if(true, x, y) → x
if(false, x, y) → y
perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) → if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

-(x0, 0)
-(s(x0), s(x1))
<=(0, x0)
<=(s(x0), 0)
<=(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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